ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.3

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.3

Very short answer type questions (1 to 4) :

Question 1.
(i) Write the order of the differential of the family of circles x² + y² = a², where a (> 0) is arbitrary constant.
(ii) Write the order of the differential equation of the family of parabolas y² = 4ax, where a is arbitrary constant.
(iii) Write the order of the differential equation of the family of parabolas y² = 4a (x – b), a, b are arbitrary constants.
(iv) Write the order of the differential equation of the family of ellipses \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, a, b are arbitrary constants.
Solution:
(i) Given eqn. of family of circles be
x² + y² = a² ………………..(1)
where a be an arbitrary constant
duff. eqn. (1) w.r.t. x; we have
2x + 2yy’ = 0 ………………..(2)
The highest ordered derivative existing in eqn.(2) be \(\frac{d y}{d x}\) so its order be 1.

(ii) Given eqn. of family of parabolas be
y² = 4ax …………………..(1)
duff. eqn. (1) w.r.t. x, we have
2yy’ = 4a
The highest ordered derivative in given diff. eqn. be y’.
∴ its order be 1.

(iii) Given y² = 4a (x – b) ;
where a, b are arbitrary constants.
2yy’ = 4a ;
again diff. both sides w.r.t. x
2[yy” + y’²] = 0
⇒ yy” + y’² = 0
The highest ordered derivative existing in given diff. eqn. be y” so its order be 2.

(iv) Given eqn. of family of ellipses be;
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 ……………(1)
Diff. eqn. (1) w.r.t. x ; we have

Here, the highest ordered derivative existing in diff. eq. (2) be \(\frac{d^2 y}{d x^2}\) and hence its order be 2.

Question 2.
(i) Form the differential equation of the . family of straight lines y = mx, where m is arbitrary constant.
(ii) Find the differential equation of the family of straight lines y = mx + c, where m, c are arbitrary constants.
Solution:
(i) Given y = mx ………………(1)
Diff. (1) w.r.t. x ; we have \(\frac{d y}{d x}\) = m ………………………(2)
Eliminating m from (1) and (2); we hae dy
y = x \(\frac{d y}{d x}\) is the required diff. eqn.

(ii) Given y = mx + c, where m, c are arbitrary constants
Differentiating w.r.t. x, we have dy
\(\frac{d y}{d x}\) = m ; again differentiating w.r.t. x,
we have \(\frac{d^2 y}{d x^2}\) = 0, be the required equation.

Question 3.
Form the differential equation of the family of concentric circles x² + y² = r², where r (> 0) is arbitrary constant.
Solution:
Given x² + y² = r² ……………(1)
Diff. both sides eqn. (1) w.r.t. x, we have
2x + 2y \(\frac{d y}{d x}\) = 0
=> x + y \(\frac{d y}{d x}\) = 0 be the required differential equation.

Question 4.
Form the differential equation not containing the arbitrary constant(s) and satisfied by the following equations :
(i) y = ax³, a is arbitrary constant
(ii) x² – y² = a², a is arbitrary constant
(iii) y = c sin^-1 x, c is arbitrary constant
(iv) y² = 4ax, a is arbitrary constant.
Solution:
(i) Given eqn. of curve be, y = ax³ ……………(1)
Diff. eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 3ax² ……………..(2)
on dividing eqn. (2) by eqn. (1); we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{3 a x^2}{a x^3}=\frac{3}{x}\)
=> x \(\frac{d y}{d x}\) = 3y be the required differential eqn.

(ii) Given eqn. of curve be x² – y² = a² ……………(1)
Diff. eqn. (1) w.r.t. x; we have dy
\(\frac{d y}{d x}=\frac{c}{\sqrt{1-x^2}}\) ………….(2)
=> x – y \(\frac{d y}{d x}\) = 0 be the reqd. differential equation.

(iii) Given y = c sin^-1 x, ………………..(1)
where c be arbitrary constant
Differentiating w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{c}{\sqrt{1-x^2}}\) ………………(2)
From (1) ;
c = \(\frac{y}{\sin ^{-1} x}\)
from (2) ;
\(\frac{d y}{d x}=\frac{y}{\sqrt{1-x^2} \sin ^{-1} x}\)
be the required diff. eqn.

(iv) Given eqn. of curve be, y² = 4ax …………..(1)
diff. eqn. (1) w.r.t. x; we have dy
2y \(\frac{d y}{d x}\) = 4a
From (1);
y² = 2xy \(\frac{d y}{d x}\)
=> 2x \(\frac{d y}{d x}\) = y be the required differential equation.

Find the differential equations by eliminating the arbitrary constants and satisfied by the following (5 to 11) equations :

Question 5.
(i) y = e^mx, m is arbitrary constant
(ii) y = k e^{tan-1 x}, k is arbitrary constant
(iii) y = tan^-1 x + ce^{– tan-1} x, c is arbitrary constant.
Solution:
(i) Given eqn. of curven be, y = e^mx ………………(1)
Diff. (1) w.r.t. x; we have dy dx dy
=> \(\frac{d y}{d x}\) = e^mx . m
=> \(\frac{d y}{d x}\) = my …………….(2) [Using (!)]
also from (1) ;
log y = mx
=> m = \(\frac{1}{x}\) log y
∴ From (2);
\(\frac{d y}{d x}=\frac{y \log y}{x}\)
=> x = y \(\frac{d y}{d x}\) the required differential equation.

(ii) Given y = k e^{tan-1 x} ……………….(1)
where k is arbitrary constant.
On differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = k e^{tan-1 x} \(\frac{1}{1+x^2}\)
=> (1 + x²) be the required diff. eqn.

(iii) Given y = tan^-1 x + c e^{– tan-1 x}
where c be arbitrary constant
Diff. eqn. (1) w.r.t. x ; we have dy_
\(\frac{d y}{d x}=\frac{1}{1+x^2}+c e^{-\tan ^{-1} x}\left(-\frac{1}{1+x^2}\right)\)
=> (1 + x²) \(\frac{d y}{d x}\) = 1 – ce^{– tan-1 x}
=> (1 + x²) \(\frac{d y}{d x}\) + y = 1 – (y – tan^-1 x) [using eqn. (1)]
=> (1 + x²) \(\frac{d y}{d x}\) + y = 1 + tan^-1 x be the required diff. eqn.

Question 6.
(i) x² + (y – b)² = 1, b is arbitrary constant
(ii) (x – a)² – y² = 1, a is arbitrary constant
(iii) x² + y² = ax³, a is arbitrary constant.
Solution:
(i) Given eqn. of curve be
x² + (y – b)² = 1 …………………(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2 (y – b) \(\frac{d y}{d x}\) = 0
=> (y – b) = \(\frac{-x}{\frac{d y}{d x}}\)
From (1);
x² + \(\frac{x^2}{\left(\frac{d y}{d x}\right)^2}\) = 1
=> x + \(\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(\frac{d y}{d x}\right)^2\) be the required differential equation.

(ii) Given eqn. of curve be,
(x – a)² – y² = 1 …………………..(1)
Diff. eqn. (1) w.r.t. x; we get
2 (x – a) – 2y \(\frac{d y}{d x}\) = 0
=> x – a = y \(\frac{d y}{d x}\)
From (1);
(y \(\frac{d y}{d x}\))² – y² = 1
y² (\(\frac{d y}{d x}\))² – y² = 1 be the required differential equation.

(iii) Given eqn. of curve be, x² + y² = ax³ ………………..(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2y \(\frac{d y}{d x}\) = 3ax² ……………………(2)
Using eqn. (1) in eqn. (2); we have
2x + 2y \(\frac{d y}{d x}\) = 3 \(\left[\frac{x^2+y^2}{x}\right]\)
⇒ 2x² + 2xy \(\frac{d y}{d x}\) = 3x² + 3y²
⇒ x² + 3y² = 2xy \(\frac{d y}{d x}\)
be the required differential equation.

Question 7.
\(\frac{x}{a}+\frac{y}{b}\) = 1, a, b are arbitrary constants. (NCERT)
Solution:
Given \(\frac{x}{a}+\frac{y}{b}\) = 1 ……………….(1)
where a and b are arbitrary constants
Diff. eqn. (1) w.r.t. x ; we get
\(\frac{1}{a}+\frac{1}{b} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{b}{a}\)
On differentiating both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 0 be the required diff. eqn.

Question 8.
(i) y² = 4a (x – b), a, b are arbitrary constants.
(ii) (y – b)² = 4 (x – a), a, b are arbitrary constants.
(iii) y² = a (b² – x²), a, bare arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be
y² = 4a (x – b) ……………(1)
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
again diff. eqn. (2) w.r.t. x ; we have
2 \(\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]\) = 0
⇒ y \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0 be the required differential equation.

(ii) Given (y – b)² = 4 (x – a)
Diff. (1) w.r.t. x ; we get
2 (y – b) y₁ = 4
again diff. both sides w.r.t. x, we get
2 [(y – b)y₂ + y₁²] = 0
Eliminating b from (2) and (3); we have
\(\frac{4}{y_1}\) . y₂ + 2 y₁² = 0
⇒ 2y₂ + y₁³ = 0 is the required differential equation.

(iiii) Given y² = a (b² – x)
Diff. eqn. (1) w.r.t. x ; we get
2yy₁ = – 2ax
⇒ \(\frac{y y_1}{x}\) = – a
diff. eqn. (2) w.r.t. x; we get
\(\frac{x\left(y y_2+y_1^2\right)-y y_1}{x^2}\) = 0
⇒ x (yy² + y¹²) – yy¹ = 0 be the reqd. diff. eqn.

Question 9.
(i) y = a cos (x + b), a, b are arbitrary constants.
(ii) y = A cos 2x + B sin 2x, A, B are arbitrary constants.
Solution:
(i) Given y = a cos (x + b) ………………….(1)
DiflF. (1) w.r.t. x ; we have
y’ = – a sin (x + b) …………………(2)
Diff. eqn. (2) w.r.t. x, we get
y” = – a cos (x + b) = -y [using (1)]
=> y” + y = 0 is the required diff. eqn.

(ii) Given y = a cos 2x + b sin 2x
where A, B are arbitrary constants.
On differentiating eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – 2A sin 2x + 2B cos 2x ax
again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – 4A cos 2x – 4B sin 2x
= – 4 (A cos 2x + B sin 2x)
=> \(\frac{d^2 y}{d x^2}\) = – 4y [using eqn. (1)]
i.e. \(\frac{d^2 y}{d x^2}\) + 4y = 0 be the required diff. eqn.

Question 10.
(i) y = A e^2x + B e^{– 2x}, A, B are arbitrary constants. (NCERT Exemplar)
(ii) y = a e^3x + b e^{– 2x}, a, b are arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be y = A e^2x + B e^{– 2x} ……………….(1)
Diff. (1) w.r.t. x; we get
\(\frac{d y}{d x}\) = 2Ae^2x – 2Be^-2x ………………(2)
Again diff. (2) w.r.t. x, we have d2 v
\(\frac{d^2 y}{d x^2}\) = 4A e^2x + 4B e^{– 2x} = 4y [Using eqn. (1)]
be the required differential equation.

(ii) Given y = A e^3x + B e^{– 2x} ………….(1)
Diff. (1) w.r.t. x ; we get
y’ = 3A e^3x – 2 B e^{– 2x} ……………….(2)
eqn. (2) – 3 × eqn. (1) ; we have
y’ – 3y = – 5B e^{– 2x} ……………(3)
Diff. (2) w.r.t. x, we get
y” – 3y’ = + 10B e^{– 2x} ………………….(4)
eqn. (4) + 2 × eqn. (3); we have
y” – 3 y’ + 2 (y’ – 3y) = 0
=> y” – y’ – 6y = 0 is the required diff. eqn.

Question 11.
(i) y = (sin^-1 x)² + A cos^-1 x + B where A, B are arbitrary constants.
(ii) y = e* (A cos x + B sin x), A, B are arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be
y = (sin^-1 x)² + A(cos^-1 x) + B
where A and B are arbitrary constants
Diff. eqn. (1) w.r.t. x; we have

(ii) Given y = e^x (A cos x + B sin x) ……………..(1)
Diff. (1) w.r.t. x ; we get
=> y’ = e^x (A cos x + B sin x) + e^x (- A sin x + B cos x)
=> y’ = y + e^x (- A sin x + B cos x) ……………….(2)
Diff. eqn. (2) w.r.t. x; we have
y” = y’ + e^x (- A sin x + B cos x) + e^x (- A cos x – B sin x)
y” = y’ + y’ – y – y
Thus, y” – 2y’ + 2y = 0
be the required differential equation.

Question 12.
Find the differential equation of the family of curves (x – h)² + (y – k)² = r² where h, k are arbitrary constants.
Solution:
Given eqn. of curve be
(x – h)² + (y – k)² = r² ………………(1)
where h and k are parameters
diff. eqn. (1) w.r.t. x; we have
2(x – h) + 2 (y – k) = 0
⇒ (x – h) + (y – k) = 0
Again diff. eqn. (2) w.r.t. x; we have
1 + (y – k) \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0
∴From (3); we have
(y – k) = – \(\frac{1+p^2}{y_2}\) ;
where p = \(\frac{d y}{d x}\) & y₂ = \(\frac{d^2 y}{d x^2}\)
From (2), we have
(x – h) = – (y – k) \(\frac{d y}{d x}=\frac{\left(1+p^2\right) p}{y_2}\)
putting the values of (x – h) and (y – k) in eqn. (1); we have
\(\frac{\left(1+p^2\right)^2 p^2}{y_2{ }^2}+\frac{\left(1+p^2\right)^2}{y_2{ }^2}\) = r²
⇒ (1 + p²)³ = r²y₂² be the reqd. eqn.

Question 13.
(i) concentric circles with centre at (1, 2).
(ii) all circles which touch they-axis at origin.
(iii) circles in the second quadrant and touching the coordinate axes.
(iv) all parabolas having their vertices at origin and foci on y-axis.
(v) all ellipses whose centres are at origin and foci on y-axis. (NCERT)
(vii) all hyperbola having foci on x-axis and centres at the origin. (NCERT)
(vii) all non-vertical lines in a plane. (NCERT Exemplar)
Solution:
(i) eqn. of all concentric circles with centre at (1, 2) and radius r is given by
(x – 1)² + (y – 2)² = r² …………………..(1)
where r be arbitrary constant.
Duff. eqn. (1) w.r.t. x ; we have
2 (x – 1) + 2(y – 2) \(\frac{d y}{d x}\) = 0
⇒ (x – 1) + (y – 2) y₁ = 0 be the required differential eqn.

(ii) We know that eqn. of family of circles touching y-axis at origin is given by
(x – r)² + y² = r² ………………….(1)
where r be the arbitrary constant.

Diff. eqn. (1) w.r.t. x ; we have
2 (x – r) + 2yy’ = 0
⇒ x – r = – yy’
and r = x + yy’
∴ From eqn. (1) ; we have
(yy’)² + y² = (x + yy’)²
⇒ (yy’)² + y² = x² + (yy’)² + 2xyy’
⇒ x² – y² + 2xyy’ = 0 is the required differential eqn.

(iii) We know that. eqn. of family of circles touching coordinate axes lies in the 2nd quadrant is given by

(x + r)² + (y – r)² = r² …………(1)
where r be thc radius of any number of family of circles and is the only one arbitrary constant.
(1) can be written as
x² + y² + 2rx – 2ry + r² = 0 ……………(2)
Diff. (2) w.r.t. x; we have
2x + 2yy’ + 2r – 2ry’ = 0
⇒ \(\frac{x+y y^{\prime}}{-1+y^{\prime}}\) = + r
Putting in eqn. (1) ; we have
\(\left[x+\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2+\left[y-\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2\)
⇒ \(\left[\frac{x y^{\prime}+y y^{\prime}}{y^{\prime}-1}\right]^2+\left[\frac{-x-y}{y^{\prime}-1}\right]^2=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2\)
⇒ y’² (x + y)² + (x + y)² = (x + yy’)²
⇒ (x + y)² (1 + y’)² = (x + yy’)² be the required diff. eqn.

(iv) equation of all parabolas having their vertices
at origin and foci on y-axis ¡s given by
x² = 4ay …………………..(1)
where a be arbitrary constant
Diff. eqn. (1) w.r.t. x ; we have
2x = 4a \(\frac{d y}{d x}\) ……………….(2)
On dividing (2) by (1) ; we have
\(\frac{2}{x}=\frac{\frac{d y}{d x}}{y}\)
⇒ x \(\frac{d y}{d x}\) = 2y be the required differential eqn.

(v) We know that eqn. of such family of ellipses is given by
\(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1 …………………(1) (a > b > 0)
where a and b are two arbitrary constants
Diff. (1) w.r.t, x ; we get
\(\frac{2 x}{b^2}+\frac{2 y y^{\prime}}{a^2}\) = 0
⇒ \(\frac{y y^{\prime}}{x}=-\frac{a^2}{b^2}\) ………………(2)
Diff. eqn. (2) w.r.t. x; we get
\(\frac{x\left(y y^{\prime \prime}+y^{\prime 2}\right)-y y^{\prime}}{x^2}\) = 0
xyy” + xy’² = yy’ is the required diff. eqn.

(vii) We know that eqn. of such family of hyperbolas is given by
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 ……………..(1) (a > 0, b > 0)
Diff. (1) w.r.t, x; we have
\(\frac{2 x}{a^2}-\frac{2 y y^{\prime}}{b^2}\) = 0
⇒ \(\frac{y y^{\prime}}{x}=\frac{b^2}{a^2}\) ………………….(2)
Diff. (2) w.r.t. x ; we have
\(\frac{x\left(y y^{\prime \prime}+y^{\prime 2}\right)-y y^{\prime}}{x^2}\) = 0
i.e. xyy” + y’² = yy’ is the required differential eqn.

(viii) We know that eqn. of any line in plane be given by
ax + by = 1
Now eqn. of any line || to y-axis be
x = constant
∴ b = 0, a ≠ 0
Thus eqn. of all non vertical lines in a plane is given by
ax + by = 1, b ≠ 0, a ∈ R
On differentiating w.r.t. x ; we have
a + b \(\frac{d y}{d x}\) = 0
again differentiating w.r,t. x ; we have
b \(\frac{d^2 y}{d x^2}\) = 0
⇒ \(\frac{d^2 y}{d x^2}\) = 0 [∵ b ≠ 0]
which is the required diff. eqn.

Question 13 (old).
(iv) all parabols having their vertices at origin and axis along positive direction of y-axis.
Solution:
Let (0, a) be the focus of any number of family of parabolas such.
∴ eqn. of such family of parabolas is given by
x² = 4ay ………………….(1)
Differential eqn. (1) w.r.t. x; we have
2x = 4ay’
⇒ y’ = \(\frac{x}{2a}\)
putting in eqn. (1); we get
x² = 4y \(\left(\frac{x}{2 y^{\prime}}\right)\)
⇒ xy’ = 2y is the required diff. eqn.

Question 14.
Form the differential equation of the family of circles having centres on y-axis and
(i) radius 3 units (NCERT)
(ii) passing through origin.
Solution:
(i) We know that eqn. of family of circle having centre on y-axis
i.e. (0, k) and radius 3 is given by
x² + (y – k)² = 3² = 9 …………..(1)
where k be arbitrary constant
Diff. (1) w.r.t. x ; we get
2x + 2 (y – k) y’ = 0
⇒ x + (y – k) y’ = 0
⇒ (y – k) = – \(\frac{x}{y^{\prime}}\)
∴ From (1);
x² + (- \(\frac{x}{y^{\prime}}\))² = 9
⇒ x² y’² = 9 is the required diff. eqn.

(ii) We know that, eqn. of circle passing through origin and centre on y-axis is
x² + y² + 2fy = 0 …………….(1)
f be any arbitrary constant.
Diff. (1) w.r.t. x ; we have
2x + 2yy’ + 2fy’ = 0
⇒ \(\frac{x+y y^{\prime}}{y^{\prime}}\) = – f
∴ From (1) ; we have
x² + y² + 2y \(\left[\frac{-x-y y^{\prime}}{y^{\prime}}\right]\) = 0
⇒ (x² + y²) y’ – 2xy – 2y²y’ = 0
⇒ (x² – y²) y’ = 2xy is the required differential equation.

Question 15.
Form the differential equation of simple harmonic motion given by x = A cos (nt + α), where n is fixed and A, α are parameters.
Solution:
Given x = A cos (nt + α) ………………(1)
where n is fixed. A, α are parameters
Diff. eqn. (1) w.r.t. t ; we have
\(\frac{d x}{d t}\) = – A sin (nt + α) . n
again differentiating w.r.t. t, we have
\(\frac{d^2 x}{d t^2}\) = – A cos (nt + α) . n² = – n²x [using eqn. (I)]
⇒ \(\frac{d^2 x}{d t^2}\) + n²x = 0, be the required diff. eqn.