ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations MCQs

Interactive ML Aggarwal Class 12 ISC Solutions engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations MCQs

Choose the correct answer from the given four options in questions (1 to 53):

Question 1.
The degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
(a) 1
(b) 2
(c) 3
(d) Not defined
Solution:
(d) Not defined

Given diff. eqn. be, \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
Here R.H.S. of given diff. eqn. cannot be expressed as polynomial in derivative so degree of given diff. eqn. be not defined.

Question 2.
The degree of the differential equation \(\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{3}{2}}=\frac{d^2 y}{d x^2}\) is
(a) 4
(b) \(\frac{3}{2}\)
(c) 2
(d) not defined
Solution:
(c) 2

Given diff. eqn. be, \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}\)
On squaring both sides ; we have
\(\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3\)
The highest order derivative present in given diff. eqn. \(\frac{d^2 y}{d x^2}\) and its power is 2.
Thus the order of given diff. eqn. be 2 and degree 2.
Clearly it is a non linear differential eqn., as it contains terms like \(\left(\frac{d^2 y}{d x^2}\right)^2\) and \(\left(\frac{d y}{d x}\right)^2\).

Question 3.
The order and the degree of the differential equation \(\left(\frac{d y}{d x}\right)^5+3 x y\left(\frac{d^3 y}{d x^3}\right)^2+y^2\left(\frac{d^2 y}{d x^2}\right)^3\) = 0 respectively, are
(a) 2, 3
(b) 3, 2
(c) 3, 5
(d) 1, 5
Solution:
(b) 3, 2

The highest ordered derivative existing in given differential eqn. be \(\frac{d^3 y}{d x^3}\) so its order be 3.
Further, given diff. eqn. can be expressed as polynomial in derivatives.
So the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 2 gives the degree of given diff. eqn.

Question 4.
The degree of the differential equation \(\left\{3-\left(\frac{d y}{d x}\right)^2\right\}^{5 / 3}=x^3\left(\frac{d^2 y}{d x^2}\right)\), is
(a) 2, 1
(b) 2, 2
(c) 2, 3
(d) 2, 5
Solution:
(c) 2, 3

Given differential eqn. be \(\left\{3-\left(\frac{d y}{d x}\right)^2\right\}^{5 / 3}=x^3\left(\frac{d^2 y}{d x^2}\right)\)
cubing on both sides ; we have
\(\left[3-\left(\frac{d y}{d x}\right)^2\right]^5=x^9\left(\frac{d^2 y}{d x^2}\right)^3\) …………….(1)
Here highest ordered derivative present in given diff. eqn (1) be \(\frac{d^2 y}{d x^2}\).
Hence, its order is 2 and also its power be 3.
∴ degree of given eqn (1) be 3.

Question 5.
The order and the degree of the differential equation \(\left(1+3 \frac{d y}{d x}\right)^{\frac{2}{3}}=4 \frac{d^3 y}{d x^3}\), respectively, are
(a) 3, \(\frac{3}{2}\)
(b) 3, 1
(c) 3, 2
(d) 3, 3
Solution:
(d) 3, 3

Given diff. eqn. be
\(\left(1+3 \frac{d y}{d x}\right)^{\frac{2}{3}}=4 \frac{d^3 y}{d x^3}\)
On cubing both sides; we have
\(\left(1+3 \frac{d y}{d x}\right)^2=\left(4 \frac{d^3 y}{d x^3}\right)^3\)
The highest ordered derivative existing in given diff. eqn. be \(\frac{d^3 y}{d x^3}\)
so its order be 3.
Further given diff. eqn. can be expressible as polynomial in derivatives.
Thus degree of given diff. eqn. be the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 3.

Question 6.
The order and degree of the differential equation \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}\) = 0
(a) 2, 4
(b) 2, 2
(c) 2, 3
(d) 2, not defined
Solution:
(a) 2, 4

Given diff. eqn be
\(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}\) = 0
⇒ \(\frac{d^2 y}{d x^2}+x^{1 / 5}=-\left(\frac{d y}{d x}\right)^{1 / 4}\)
⇒ \(\left[\frac{d^2 y}{d x^2}+x^{y / 5}\right]^4=+\frac{d y}{d x}\)
which is polynomial in derivatives.
Also the highest ordered derivative existing in given diff. eqn be and its power be 4.
Thus the order of given differential eqn. be 2 and its degree be 4.

Question 7.
The order of the differential equation representing the family of curves y = ax + a³
(a) 1, 1
(b) 1, 3
(c) 1, 2
(d) 2, 3
Solution:
(b) 1, 3

y = ax + a³
∴ From given eqn ; we have
y = x \(\frac{d y}{d x}\) + (\(\frac{d y}{d x}\))³
which is diff. eqn. of order 1 and degree 3.

Question 8.
The order of the differential equation of the family of circles of radius r.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

Eqn. of families of circle with centre (h, k)
and radius r is given by
(x – a)² + (y – b)² = r² ……………..(1)
Diff. (1) w.r.t. x,. we have
2 (x – a) + 2 (y – b) y₁ = 0
⇒ (x – a) + (y – b) y₁ = 0 ……………….(2)
Diff. eqn (2) w.r.t. x; we have
1 + (y + b)y₂ + y₁² = 0
⇒ (y – b) = – \(\frac{\left(1+y_1^2\right)}{y_2}\)
∴ From (2); we have
x – a = – (y – b) y₁
= \(\frac{\left(1+y_1^2\right) y_1}{y_2}\)
∴ From eqn (1) ; we have
\(\frac{\left(1+y_1^2\right)^2 y_1^2}{y_2^2}+\frac{\left(1+y_1^2\right)^2}{y_2^2}\) = r²
clearly required order of duff. eqn. be 2.

Question 9.
Which of the following is a second order differential equation?
(a) (y’)² + x = y²
(b) y’ = y²
(c) y’y” + y = sin x
(d) y”’ + (y”)² + y = o.
Solution:
(c) y’y” + y = sin x

Clearly the highest ordered derivative existing in given duff. eqn. y’y” + y = sin x be y”. so its order be 2.

Question 10.
The differential equation of the family
of curves x² + y² – 2ay = 0, where a is arbitrary constant, is
(a) (x² – y²) \(\frac{d y}{d x}\) = 2xy
(b) 2 (x² + y²) \(\frac{d y}{d x}\) = xy
(c) 2 (x² – y²) \(\frac{d y}{d x}\) = xy
(d) (x² + y²) \(\frac{d y}{d x}\) = 2xy
Solution:
(a) (x² – y²) \(\frac{d y}{d x}\) = 2xy

Given diff. eqn. be, x² + y² – 2ay = 0 ……………(1)
Diff. eqn. (1) both sides w.r.t. x; we have
2x + 2y – 2a \(\frac{d y}{d x}\) = 0
⇒ (x + y \(\frac{d y}{d x}\)) = a \(\frac{d y}{d x}\)
∴ from (1) ; we have
x² + y² – 2y \(\frac{\left(x+y \frac{d y}{d x}\right)}{\frac{d y}{d x}}\) = 0
⇒ (x² + y²) \(\frac{d y}{d x}\) – 2xy – 2y² \(\frac{d y}{d x}\) = 0
⇒ (x² – y²) \(\frac{d y}{d x}\) = 2xy
which is the required differential equation.

Question 11.
The differential equation of the family of curves y² = 4a (x + a), where a is arbitrary constant, is
(a) y² = 4 \(\frac{d y}{d x}\) (x + \(\frac{d y}{d x}\))
(b) y \(\frac{d y}{d x}\) = 2a
(c) y \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0
(d) 2x \(\frac{d y}{d x}\) + y (\(\frac{d y}{d x}\))² = y
Solution:
(d) 2x \(\frac{d y}{d x}\) + y (\(\frac{d y}{d x}\))² = y

Given eq. be curve be
y² = 4a (x + a)
where a be an arbitrary constant.
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ a = \(\frac{y}{2}\) \(\frac{d y}{d x}\)
∴ from (1) ; we have
y² = 2y \(\frac{d y}{d x}\) [x + \(\frac{y}{2} \frac{d y}{d x}\)]
⇒ y² = 2xy \(\frac{d y}{d x}\) + y² (\(\frac{d y}{d x}\))²

Question 12.
The differential equation for which y = a cos x + b sin x is a solution, is [NCERT Exemplar]
(a) \(\frac{d^2 y}{d x^2}\) + y = 0
(b) \(\frac{d^2 y}{d x^2}\) – y = 0
(c) \(\frac{d^2 y}{d x^2}\) + (a + b) y = 0
(d) \(\frac{d^2 y}{d x^2}\) + (a – b) y = 0
Solution:
(a) \(\frac{d^2 y}{d x^2}\) + y = 0

Given eqn. of solution be,
y = a cos x + b sin x ………………..(1)
Diff. eqn. (1) both sides w.r.t. x; we have
\(\frac{d y}{d x}\) = – a sin x + b cos x
Again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – a cos x – b sin x = – y
[using eqn. (1)]
⇒ \(\frac{d^2 y}{d x^2}\) + y = 0
which is the required differential equation.

Question 13.
y = ae^mx + be^-mx, where a, b are arbitrary constants satisfies which of the following differential equation?
(a) \(\frac{d y}{d x}\) + my = 0
(b) \(\frac{d y}{d x}\) – my = 0
(c) \(\frac{d^2 y}{d x^2}\) – m²y = 0
(d) \(\frac{d^2 y}{d x^2}\) + m²y = 0
Solution:
(c) \(\frac{d^2 y}{d x^2}\) – m²y = 0

Given y = ae^mx + be^-mx ………………(1)
where a and b are arbitrary constants
diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = am e^mx – bm e^{– mx}
∴ \(\frac{d^2 y}{d x^2}\) = am² e^mx + bm² e^{– mx}
= m² [ae^mx + be^{– mx}]
= m² y [using (1)]

Question 14.
The differential equation obtained on eliminating A and B from y A cos ωt + B sin ωt, is
B sin mt, is
(a) y” + y’ = 0
(b) y” – ω²y = 0
(c) y” = – ω²y
(d) y” + y = 0
Solution:
(c) y” = – ω²y

Given y = A cos ωt + B sin ωt …………….(1)
y’ = – Aω sin ωt + Bω cos ωt
∴ y”= – Aω² cos ωt – Bω² sin ωt
= – ω² [A cos ωt + B sin ωt]
= – ω²y

Question 15.
The differential equation for the family of curves sin^-1 x + sin^-1 y = C, where C Is arbitrary constant, is
(a) \(\sqrt{1-x^2} d x+\sqrt{1-y^2} d y\) = 0
(b) \(\sqrt{1-x^2} d x-\sqrt{1-y^2} d y\) = 0
(c) \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0
(d) \(\sqrt{1-x^2} d y-\sqrt{1-y^2} d x\) = 0
Solution:
(c) \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0

Given sin^-1 x + sin^-1 y = C
where C be an arbitrary constant
Diff. eqn. (1) w.r.t. x;
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d x}{\sqrt{1-x^2}}+\frac{d y}{\sqrt{1-y^2}}\) = 0
⇒ \(\sqrt{1-y^2} d x+\sqrt{1-x^2} d y\) = 0

Question 16.
tan^-1 x + tan^-1 y = C, where C Is a parameter, is the general solution of the differential equation:
(a) \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
(b) \(\frac{d y}{d x}=\frac{1+x^2}{1+y^2}\)
(c) (1+ x²) dy + (1 + y²) dx = 0
(d) (1 + x²) dy + (1 + y²) dy = 0
Solution:
(c) (1+ x²) dy + (1 + y²) dx = 0

Given tan^-1 x + tan^-1 y = C ……………………(1)
diff. eqn. (1) w.r.t. x ; we have
\(\frac{1}{1+x^2}+\frac{1}{1+y^2} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{1+y^2}{1+x^2}\)

Question 17.
The differential equation representing the family of curves y² = 2c (x + √c) where c (> 0) is a parameter, is of
(a) order 2, degree 2
(b) order 1, degree 3
(c) order 1, degree 1
(d) order 1. degree 2
Solution:
(b) order 1, degree 3

Given y² = 2c (x + √c)
where c be an arbitrary constant
diff. eqn. (1) w.r.t. x ; we have
2yy’ = 2c
∴ from (1) ;
y² = 2yy’ [x + \(\)]
⇒ y = 2xy’ + 2y’ \(\sqrt{y y^{\prime}}\)
⇒ (y – 2xy’)² = 4y’² (yy’) = 4yy’³
Here the highest ordered derivative existing in given diff. eqn. be y’ so its order be 1.
Clearly the given diff. eqn. can be expressed as polynomial in derivatives.
Thus, the degree of given diff. eqn. be the highest exponent of y’ which is 3.

Question 18.
Which of the following ¡s not a homogeneous function of x and y?
(a) x² + 2y
(b) 2x – y
(c) cos² \(\left(\frac{y}{x}\right)+\frac{y}{x}\)
(d) sin x – cos y
Solution:
(d) sin x – cos y

x² + 2xy = x² [1 + 2(\(\frac{y}{x}\))]
= x² Φ (\(\frac{y}{x}\))
2 – xy = x [2 – \(\frac{y}{x}\)]
= x g (\(\frac{y}{x}\)) ;
cos² (\(\frac{y}{x}\)) + \(\frac{y}{x}\) = h (\(\frac{y}{x}\))
All of these functions are homogeneous in x and y.
Further sin x – sin y cant be expresed in the form of f (\(\frac{y}{x}\)).

Question 19.
Which of the following s a homogeneous differential equation?
(a) (2x – 3y + 5) dy – (3x + 2y – 4) dx = 0
(b) xy dx – (x³ – y³) dy = 0
(c) y² dx – (x² – xy + y²) dy = 0
(d) (y³ + 2xy) dx + 3x² dy = 0
Solution:
(c) y² dx – (x² – xy + y²) dy = 0

Clearly in option (c);
The given diff. eqn. can be written as;
\(\frac{d y}{d x}=\frac{y^2}{x^2-x y+y^2}\)
= \(\frac{\left(\frac{y}{x}\right)^2}{1-\frac{y}{x}+\left(\frac{y}{x}\right)^2}\)
= \(\phi\left(\frac{y}{x}\right)\)
which is a homogeneous function of degroe 0.

Question 20.
The number of solutions of \(\frac{d y}{d x}=\frac{y+1}{x-1}\), when y(1) = 2 is
(a) none
(b) one
(c) two
(d) infinite
Solution:
(b) one

Given \(\frac{d y}{d x}=\frac{y+1}{x-1}\) ……………..(1)
after variable separation, we have
\(\frac{d y}{y+1}=\frac{d x}{x-1}\) ;
on integrating

Question 21.
Integrating factor of the differential equation x \(\frac{d y}{d x}\) + y = 3x² is
(a) x
(b) log x
(c) \(\frac{1}{x}\)
(d) none of these
Solution:
(a) x

Given diff. eqn. can be written as;
\(\frac{d y}{d x}+\frac{y}{x}\) = 3x
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = \(\frac{1}{x}\) and Q = 3x
∴ I.F = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x

Question 22.
Integrating factor of x \(\frac{d y}{d x}\) – y = x⁴ -3x is
(a) x
(b) log x
(c) \(\frac{1}{x}\)
(d) – x
Solution:
Given diff. eqn. can be written as ;
\(\frac{d y}{d x}\) – \(\frac{y}{x}\) = 4x³ -3
which is of the form \(\frac{d y}{d x}\) + Py = Q
where P = – \(\frac{1}{x}\) and Q = x³ – 3
∴ I.F = e^{∫ P dx}
= \(\int_e-\frac{1}{x} d x\)
= e^{– log x}
= e ^{log x-1}
= x^-1
= \(\frac{1}{x}\)

Question 23.
Integrating factor of the differential equation cos x \(\frac{d y}{d x}\) + y sin x = 1, is
(a) sin x
(b) sec x
(c) tan x
(d) cos x
Solution:
(b) sec x

Given differential equation be,
cos x \(\frac{d y}{d x}\) + y sin x = 1
⇒ \(\frac{d y}{d x}+y \frac{\sin x}{\cos x}\) = sec x
which is L.D.E. of the form
\(\frac{d y}{d x}\) + Py = Q
where P = tan x
and Q = sec x
∴ I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= e^{– log cos x}
= \(e^{\log \frac{1}{\cos x}}\)
= \(\frac{1}{\cos x}\)
= sec x

Question 24.
Integratlng factor of the differential equation (1 – x²) \(\frac{d y}{d x}\) – xy = 1 is
(a) – x
(b) \(\frac{x}{1+x^2}\)
(c) \(\sqrt{1-x^2}\)
(d) \(\frac{1}{2}\) log(1 – x²)
Solution:
(c) \(\sqrt{1-x^2}\)

Given diff. eqn. can be written as;
\(\frac{d y}{d x}-\frac{x}{1-x^2} y=\frac{1}{\left(1-x^2\right)}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q

Question 25.
The integrating factor of the differential equation (x log x) \(\frac{d y}{d x}\) + y = 2 log x, given by
(a) log (log x)
(b) e^x
(c) log x
(d) x
Solution:
(c) log x

Given diff. eqn. be,
(x log x) \(\frac{d y}{d x}\) + y = 2 log x
which is linear differential eqn.
and is of the form \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x \log x}\)
and Q = \(\frac{2}{x}\)
Now I.F = e^{∫ P dx}
= \(e^{\int \frac{1 / x}{\log x} d x}\)
= e^{log (log x)}
= log x

Question 26.
Integrating factor of the differential equation (1 – y²) \(\frac{d x}{d y}\) + xy = ay is
(a) \(\frac{1}{1-y^2}\)
(b) \(\frac{1}{\sqrt{y^2-1}}\)
(c) \(\frac{1}{y^2-1}\)
(d) \(\frac{1}{\sqrt{1-y^2}}\)
Solution:
(d) \(\frac{1}{\sqrt{1-y^2}}\)

Given diff. eqn. can be written as
\(\frac{d x}{d y}+\left(\frac{y}{1-y^2}\right) x=\frac{d y}{1-y^2}\)
which is L.D.E in x and is of the form
\(\frac{d x}{d y}\) + Py = Q

Question 27.
Integrating factor of the differential equation \(\frac{d y}{d x}\) + y = \(\frac{x^3+y}{x}\) is
(a) xe^x
(b) \(\frac{x}{e^x}\)
(c) \(\frac{e^x}{x}\)
(d) e^x
Solution:
(c) \(\frac{e^x}{x}\)

Given diff eqn. can be written as ;
\(\frac{d y}{d x}+\left(1-\frac{1}{x}\right) y\) = x²
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = 1 – \(\frac{1}{x}\)
and Q = x²
∴ I.F. = e^{∫ P dx}
= \(e^{\int\left(1-\frac{1}{x}\right) d x}\)
= e^{x – log x}
= e^x e^{log x-1}
= e^x \(\frac{1}{x}\)

Question 28.
Solution of the differential \(\frac{d x}{x}\) + \(\frac{d y}{y}\) = 0 is
(a) \(\frac{1}{x}+\frac{1}{y}\) = C
(b) xy = C
(c) log x log y = C
(d) x + y = C
Solution:
(b) xy = C

Given diff eqn. can be written as ;
\(\frac{d x}{x}\) + \(\frac{d y}{y}\) = 0
On integrating both sides ; we have
\(\int \frac{d x}{x}+\int \frac{d y}{y}\) = log C
⇒ log |x| + log |y| = log C
⇒ log |xy| = log C
⇒ xy = ± C = A

Question 29.
The solution of the differential equation \(\frac{x d y}{d x}\) + 2y = x² is
(a) y = \(\frac{x^2}{4}\) + C
(b) y = \(\frac{x^2+C}{4 x^2}\)
(c) y = \(\frac{x^2+C}{x^2}\)
(d) y = \(\frac{x^4+C}{4 x^2}\)
Solution:
(d) y = \(\frac{x^4+C}{4 x^2}\)

Given differential equation can be written as;
\(\frac{d y}{d x}+\frac{2}{x} y\) + 2y = x²
which is first order L.D.E. in y and is of the form
\(\frac{x d y}{d x}\) + Py = Q ;
where P = \(\frac{2}{x}\) and Q = x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{2}{x} d x}\)
= e^{2 log |x|}
= e^{log |x|2}
= x²
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . x² = ∫ x . x² dx + C
⇒ x²y = \(\frac{x^4}{4}\) + C
⇒ y = \(\frac{x^4+C}{4 x^2}\) be the required solution.

Question 30.
Solution of the differential equation tan y sec² x dx + tan x sec² y dy is
(a) tan x+ tany = C
(b) tan x – tan y = C
(c) tan x tany = C
(d)tan x = C tan y
Solution:
(c) tan x tany = C

Given diff. eqn. can be written as ;
\(\frac{\sec ^2 x d x}{\tan x}+\frac{\sec ^2 y d y}{\tan y}\) = 0
On integrating ; we have
\(\int \frac{\sec ^2 x d x}{\tan x}+\int \frac{\sec ^2 y d y}{\tan y}\) = log C
⇒ log |tan x| + log |tan y| = log C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
⇒ log |tan x tan y| = log C
⇒ tan x tan y = ± C = A

Question 31.
The solution of the differential equation \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\) is
(a) y = tan^-1 x + C
(b) x = tan^-1 y + C
(c) tan (xy) = C
(d) y – x = C (1 + xy)
Solution:
(d) y – x = C (1 + xy)

Given \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
After variable separation, we have
\(\frac{d y}{1+y^2}=\frac{d x}{1+x^2}\) ;
on integrating
\(\int \frac{d y}{1+y^2}-\int \frac{d x}{1+x^2}\) = A
⇒ tan^-1 y – tan^-1 x = A
⇒ tan^-1 \(\left(\frac{y-x}{1+x y}\right)\) = A
⇒ y – x = C (1 + xy) ; where tan A = C

Question 32.
The solution of th differential equation \(\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}\) is
(a) sin^-1 y – sin^-1 x = C
(b) sin^-1 y + sin^-1 x = C
(c) sin^-1 (xy) = C
(d) none of these
Solution:
(a) sin^-1 y – sin^-1 x = C

After variable separation, we have
\(\frac{d y}{\sqrt{1-y^2}}=\frac{d x}{\sqrt{1-x^2}}\) ;
On integrating
\(\int \frac{d y}{\sqrt{1-y^2}}=\int \frac{d x}{\sqrt{1-x^2}}+\mathrm{C}\)
⇒ sin^-1 y – sin^-1 x = C

Question 33.
The general solution of e^x cos y dx – e^x sin y dy = 0 is
(a) e^x sin y = C
(b) e^x cos y = C
(c) e^x = C cos y
(d) e^x = C sin y
Solution:
(b) e^x cos y = C

Given diff. eqn. can be written as ;
e^x [dx – \(\frac{\sin y}{\cos y}\) dy] = 0
⇒ dx – \(\frac{\sin y}{\cos y}\) dy = 0
On integrating both sides ; we have
∫ dx = ∫ \(\frac{\sin y}{\cos y}\) dy + C
⇒ x = – log |cos y| + C
⇒ log |cos y| = C – x
⇒ |cos y| = e^{C – x}
= Ae^-x
⇒ e^x cos y = ± A = B

Question 34.
The solution of the equation (2y – 1) dx – (2x + 3) dy = 0 is
(a) \(\frac{2 x-1}{2 y+3}\) = C
(b) \(\frac{2 y+1}{2 x-3}\) = C
(c) \(\frac{2 x+3}{2 y-1}\) = C
(d) \(\frac{2 x-1}{2 y-3}\) = C
Solution:
(c) \(\frac{2 x+3}{2 y-1}\) = C

After variable separation, we have
\(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) = 0 ;
On integrating

Question 35.
The general solution of \(\frac{d y}{d x}\) + y tan x = sec x is
(a) y sec x tan x + C
(b) y tan x = sec x + C
(c) x sec x = tan y + C
(d) none of these
Solution:
(a) y sec x tan x + C

Given diff. eqn. be
\(\frac{d y}{d x}\) + y tan x = sec x
which is linear diff. eqn. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = tan x ; Q = sec x
∴ I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= \(e^{\int \frac{\sin x d x}{\cos x}}\)
= e^{– log |cos x|}
= e^{log |cos x|-1}
= \(\frac{1}{cos x}\)
= sec x
and general solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . sec x = ∫ sec² x dx + C
= tan x + C

Question 36.
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}\) is
(a) y log (1 + x²) = tan^-1 x + C
(b) y (1 + x²) = tan^-1 x + C
(c) y (1 + x²) sin^-1 x + C
(d) \(\frac{y}{1+x^2}\) = tan^-1 x + C
Solution:
(b) y (1 + x²) = tan^-1 x + C

Given diff. eqn. be,
\(\frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{1}{\left(1+x^2\right)^2}\)
which is L.D.E my and is of the form
\(\frac{d y}{d x}\) + Py = Q;
where P = \(\frac{2 x}{1+x^2}\)
and Q = \(\frac{1}{\left(1+x^2\right)^2}\)
Here I.F = e^{∫ P dx}
= \(e^{\int \frac{2 x}{1+x^2} d x}\)
= e^{log (1 + x2)}
= 1 + x²
and general solution be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . (1 + x²) = ∫ \(\frac{1}{\left(1+x^2\right)^2}\) (1 + x²) dx + C
= tan^-1 x + C

Question 37.
The general solution of the differential equation \(\frac{d y}{d x}\) + y cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C
Solution:
(d) y sin x = x + C

Given \(\frac{d y}{d x}\) + y cot x = cosec x
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = cot x & Q = cosec x
∴ I.F. = e^{∫ P dx}
= e^{∫ cot x dx}
= e^{log sin x}
= sin x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
y . sin x = ∫ cosec x . sin x dx + c
= x + c

Question 38.
The solution of the differential equation \(\frac{d y}{d}-\frac{y(x+1)}{x}\) = 0 is given by
(a) y = xe^{x + c}
(b) x= ye^x
(c) y = x + C
(d) xy = e^x + C
Solution:
(a) y = xe^{x + c}

Given diff. eqn be \(\frac{d y}{d}- y \frac{(x+1)}{x}\) = 0
⇒ \(\frac{1}{y} d y-\left(\frac{x+1}{x}\right) d x\) = 0 ;
on integrating
⇒ log y – (x + log x) = C
⇒ log \(\frac{y}{x}\) = x + C
⇒ \(\frac{y}{x}\) = x + C
⇒ \(\frac{y}{x}\) = e^{x + C}
⇒ y = xe^{x + C} be the required solution.

Question 39.
The general solution of the differential equation \(\frac{d y}{d x}\) = e^{x + y}, is
(a) e^x + e^-y = C
(b) e^x + e^y = C
(c) e^-x + e^y = C
(d) e^-x + e^-y = C
Solution:
(a) e^x + e^-y = C

Given diff. eqn be,
\(\frac{d y}{d x}\) = e^{x + y}
⇒ \(\frac{d y}{d x}\) = e^-x . e^y
⇒ e^-y dy = e^x dx
on integrating; we have
e^-y = e^x + c
⇒ e^-y + e^-x = A be the required solution.

Question 40.
General solution of the differential equation(1 + x) y dx + (1 – y) x dy = 0 is
(a) log xy + x – y = C
(b) log xy – x + y = C
(c) log (\(\frac{x}{y}\)) + x + y = C
(d) log (\(\frac{x}{y}\)) – x + y = C
Solution:
(a) log xy + x – y = C

Given differential eqn. can be written s;
\(\frac{(1+x)}{x} d x+\frac{(1-y)}{y} d y\) = 0
On integrating both sides ; we have
\(\int\left(\frac{1}{x}+1\right) d x+\int\left(\frac{1}{y}-1\right) d y\) = C
⇒ log |x| + x + log |y| – y = C
⇒ log |xy| + x – y = C

Question 41.
Solution of the differential equation \(\frac{d y}{d x}\) = e^{x – y} + x² e^-y is
(a) y = e^{x – y} – x² e^-y + C
(b) e^x + e^y = \(\frac{x^3}{3}\) + C
(c) e^y – e^x = \(\frac{x^3}{3}\) + C
(d) e^x – e^y = \(\frac{x^3}{3}\) + C
Solution:
(c) e^y – e^x = \(\frac{x^3}{3}\) + C

Given \(\frac{d y}{d x}\) = e^{x – y} + x² e^-y
= e^-y (e^x + x²)
On integrating both sides;
⇒ ∫ e^y dy = ∫ (e^x + x²) dx + C
⇒ e^y = e^x + \(\frac{x^3}{3}\) + C

Question 42.
General solution of the differential equation \(\frac{d y}{d x}\) = \(e^{\frac{x^2}{2}}\) + xy is
(a) Y = (C – x) \(e^{\frac{x^2}{2}}\)
(b) y = (x + C) \(e^{\frac{x^2}{2}}\)
(c) y = C\(e^{-\frac{x^2}{2}}\)
(d) y = C
Solution:
(b) y = (x + C) \(e^{\frac{x^2}{2}}\)

Given diff. eqn. be,
\(\frac{d y}{d x}\) = \(e^{\frac{x^2}{2}}\) + xy
⇒ \(\frac{d y}{d x}\) – xy = \(e^{\frac{x^2}{2}}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – x ; Q = \(e^{\frac{x^2}{2}}\)
∴ I.F. = e^{∫ P dx}
= e^{∫ – x dx}
= \(e^{-\frac{x^2}{2}}\)
and general solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y\(e^{-\frac{x^2}{2}}\) = ∫ \(\int e^{\frac{x^2}{2}-\frac{x^2}{2}}\) dx + C
= x + C
⇒ y = (x + C) \(e^{\frac{x^2}{2}}\)

Question 43.
General solution of the differential equation \(\frac{d y}{d x}\) = 2x e^{x2 – y} is
(a) e^{– y} + C
(b) e^y = e^x2 + C
(c) e^{x2 – y} = C
(d) e^{x2 + y} = C
Solution:
(b) e^y = e^x2 + C

Given \(\frac{d y}{d x}\) = 2x e^{x2 – y}
= 2x e^x2 e^-y
⇒ e^y dy = 2x e^x2 dx
[after variable separation]
On integrating both sides we have
∫ e^y dy = ∫ 2x e^x2 dx + C
⇒ e^y = ∫ e^t dt + C
[where x² = t
⇒ 2x dx = dt]
⇒ e^y = e^t + C
⇒ e^y = e^x2 + C

Question 44.
General solution of the differential equation (e^x + 1) y dy = (y + 1) e^x dx is
(a) y + 1 = C (e^x + 1)
(b) y + 1 = e^x + 1 + C
(c) y = log(C (y + 1) (e^x + 1))
(d) none of these
Solution:
(c) y = log(C (y + 1) (e^x + 1))

Given (e^x + 1) y dy = (y + 1) e^x dx
After variable separation, we have

[where e^x = t
= e^x dx = dt]
⇒ y – log |y + 1| = log |t + 1| + cos C
⇒ y – log |y + 1| = log |e^x + 1| + log c
⇒ y = log {|y + ||e^x + 1| C}
⇒ y = log (C (y + 1) (e^x + 1))

Question 45.
General solution of the differential equation x \(\frac{d y}{d x}\) + y = e^x is
(a) y = \(\frac{e^x}{x}+\frac{C}{x}\)
(b) y = x e^x + Cx
(c) y = xe^x + C
(d) x = \(\frac{e^y}{y}+\frac{c}{y}\)
Solution:
Given diff. eqn. can be written as
\(\frac{d y}{d x}+\frac{y}{x}=\frac{e^x}{x}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = \(\frac{1}{x}\)
and Q = \(\frac{e^x}{x}\)
I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x} = x
and general solution be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
y . x = ∫ \(\frac{e^x}{x}\) . x dx + C
= e^x + C
⇒ y = \(\frac{e^x}{x}+\frac{c}{x}\)

Question 46.
Solution of the differential equation \(\frac{d y}{d x}\) + ay = e^mx is
(a) (a + m) y = e^mx + Ce^{– ax}
(b) y = e^mx + Ce^-ax
(c) (a + m) y = e^mx + C
(d) ye^ax = me^mx + C
Solution:
(a) (a + m) y = e^mx + Ce^{– ax}

Given diff. eqn. be,
\(\frac{d y}{d x}\) + ay = e^mx
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = a ; Q = e^mx
∴ I.F. = e^{∫ P dx}
= e^{∫ a dx}
= e^ax
and soln, be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + C
y . e^ax = ∫ e^mx . e^ax dx + C
= ∫ e^{(m + a) x} dx + C
= \(\frac{e^{(m+a) x}}{m+a}\) + C
⇒ (m + a) y = e^mx + C’ e ^ax

Question 47.
Solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}\) = sin x is
(a) x (y + cos x) = sin x + C
(b) x(y – cos x) = sin x + C
(c)x (y + cos x) = cos x + C
(d) none of these
Solution:
(a) x (y + cos x) = sin x + C

Given diff. eqn be,
\(\frac{d y}{d x}+\frac{y}{x}\) = sin x
which is L.D.E. of the from \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x}\) and Q = sin x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x
and solution is given by
y . e^{∫ P dx} = Q . ∫ e^{∫ P dx} dx + c
⇒ y.x = ∫ sin x . x dx + c
⇒ xy = [- x cos x + sin x] + c
⇒ x (y + cos x) = sin x + c
which is the reqd. solution.

Question 48.
General solution of the differential equation cos x sin y dx + sin x cosy dy = 0 is
(a) sin x = C sin y
(b) sin x siny = C
(e) sin x + sin.y C
(d) cos x cosy = C
Solution:
(b) sin x siny = C

Given diff. eqn. can be written as;
\(\frac{\cos x d x}{\sin x}+\frac{\cos y d y}{\sin y}\) = 0
On integrating both sides ; we have
\(\int \frac{\cos x}{\sin x} d x+\int \frac{\cos y}{\sin y} d y\) = log C
⇒ log |sin x| + log |sin y| = log C
⇒ log |sin x sin y| = log C
⇒ sin x sin y = ± C = A

Question 49.
The solution of the differential equation \(\frac{d y}{d x}\) = y = e^-x, y(0) = 0, is
(a) y = e^x (x – 1)
(b) y = xe^-x
(c) y = xe^-x + 1
(d) y = (x + 1) e^-x
Solution:
(b) y = xe^-x

Given diff. eqn. be,
\(\frac{d y}{d x}\) + y = e^-x
which is L.Ð.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
where P = 1 and Q = e^-x
∴ I.F = e^{∫ P dx}
= e^{∫ 1 dx}
= e^x
and soln. be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y e^x = ∫ e^{– x} .e^x dx + C
= ∫ 1 dx + C
= x + C ………………(1)
Given y(0) = 0 i.e.
when x = 0
⇒ y = 0
∴ from (1) ;
0 = (0 + C)
⇒ C = 0
Thus eqn. (1) becomes;
y = x e^{– x}

Question 50.
The solution of the differential equation \(\frac{d y}{d x}\) – y = 1, y(0) = 1, is
(a) xy = e^{– x}
(b) xy = – e^{– x}
(c) xy = – 1
(d) y = 2 e^x – 1
Solution:
(d) y = 2 e^x – 1

Given diff. eqn. be
\(\frac{d y}{d x}\) – y = 1
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – 1 and Q = 1
∴ I.F = e^{∫ P dx}
= e^{– ∫ dx}
= e^{– x}
and general soln, be given by
⇒ y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ ye^{– x} = ∫ 1 . e^{– x}dx + C
⇒ ye^{– x} = \(\frac{e^{-x}}{-1}\) + C
⇒ y = – 1 + Ce^x …………………(1)
Given y (0) = 1
i.e. When x = 0 ⇒ y = 1
⇒ 1 = – 1 + C
⇒ C = 2
from (1) ;
y = – 1 + 2e^x

Question 51.
Solution of the differential equation x dy -y dx = 0 represents a family of
(a) rectangular hyperbolas
(b) parabolas with vertex at origin
(c) straight lines passing through origin
(d) circles whose centre is at origin
Solution:
(c) straight lines passing through origin

Given diff. eqn. can be written as:
\(\frac{d y}{y}-\frac{d x}{x}\) = 0 ;
On integrating ; we have
\(\int \frac{d y}{y}-\int \frac{d x}{x}\) = log C
⇒ log y – log x = log C
⇒ log \(\frac{y}{x}\) = log C
⇒ y = Cx
which represents a family of straight lines passing through origin.

Question 52.
Solution of the differential equation y \(\frac{d y}{d x}\) + x = 0 represents a family of
(a) hyperbola
(b) parabolas
(c) ellipses
(d) circles
Solution:
(d) circles

Given diff. eqn. can be written as;
y dy + x dy = 0 ;
on integrating ; we have
∫ y dy + ∫ x dx = 1
⇒ \(\frac{y^2}{2}+\frac{x^2}{2}=\frac{C}{2}\)
⇒ x² + y² = C
which represents a family of circles.

Question 53.
The solution of the differential equation 2x \(\frac{d y}{d x}\) – y = 3 represents
(a) circles
(b) straight lines
(c) ellipses
(d) parabolas
Solution:
(d) parabolas

Given diff. eqn be, 2x \(\frac{d y}{d x}\) – y = 3
⇒ \(\frac{d y}{d x}-\frac{y}{2 x}=\frac{3}{2 x}\)
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q

on squaring; we have
(y + 3)² = c²x,
which represents a right handed parabolas with vertex (0, – 3).