OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Ex 17(f)

Students can track their progress and improvement through regular use of ISC Class 12 OP Malhotra Solutions Chapter 17 Differential Equations Ex 17(f).

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(f)

Solve the following equations :

Question 1.
Find the integral factor of the following differential equations :
(i) \(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}\) = 1
(ii) (1 + y²) + (2xy – cot y)\(\frac { dy }{ dx }\) = 0
Solution:
(i) Given diff. eqn. can be written as
\(\frac{d y}{d x}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\)
⇒ \(\frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\) … (1)
which is linear differential eqn. of first order
On comparing eqn. (1) with \(\frac { dy }{ dx }\) + Py = Q

which is linear diff. eqn. in x of first order On comparing eqn. (1) with \(\frac { dy }{ dx }\) + Px = Q

Solve the following equations :

Question 2.
\(\frac { dy }{ dx }\) + y = e^-x
Solution:
Given diff. eqn. be \(\frac { dy }{ dx }\) + y = e^-x
which is L.D.E of first order in y.
On comparing with \(\frac { dy }{ dx }\) + Py = Q dx
Here, P = 1 ; Q = e^-x
∴ LF. = \(e^{\int P d x}=e^{\int d x}\) e^x
and solution is given by
\(y \cdot e^{\int P d x}=\int\left(Q \cdot e^{\int P d x}\right)\) dx + c
⇒ y e^x = \(\int e^{-x} \cdot e^x\) dx + c = x + c
⇒ y = (x + c) e^-x
which is the required solution.

Question 3.
x\(\frac { dy }{ dx }\) – ay = x-y2 +1
Solution:
Given diff. eqn. can be written as,
\(\frac{d y}{d x}-\frac{a}{x} y=\frac{x+1}{x}\)
which is L.D.E in y of first order.
On comparing with \(\frac { dy }{ dx }\) + Py = Q

Question 4.
(2x – 10y³)\(\frac { dy }{ dx }\) + y = 0 dx
Solution:

Question 5.
\(\frac { dy }{ dx }\) + 2y = 6e^x
Solution:
Given, \(\frac { dy }{ dx }\) + 2y = 6e^x
which is L.D.E in y of first order
On comparing eqn. (1) with \(\frac { dy }{ dx }\) + Py = Q

be the required solution.

Question 6.
(i) x\(\frac { dy }{ dx }\) – y = x²
(ii) 2x\(\frac { dy }{ dx }\) + y = 6x³
Solution:
(i) Given diff. eqn. can be
\(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) = x
which is L.D.E in y of first order
On comparing with \(\frac { dy }{ dx }\) + Py = Q dx
Here P = – \(\frac { 1 }{ x }\) and Q = x
∴ I.F = \(e^{\int \mathrm{P} d x}=e^{\int-\frac{1}{x} d x}\)
= \(e^{-\log x}=e^{\log x^{-1}}=\frac{1}{x}\)
and solution is given by
\(y \cdot e^{\int P d x}=\int \mathrm{Q} \cdot e^{\int P d x} d x+c\)
⇒ \(y \cdot \frac{1}{x}=\int x \cdot \frac{1}{x} \)dx + c = x + c
⇒ y = x (x + c)
which is the required solution.

(ii) Given diff. eqn. can be written as
\(\frac{d y}{d x}+\frac{y}{2 x}=3 x^2\)
which is L.D.E my of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = \(\frac { 1 }{ 2x }\) ; Q = 3x²

which is the required solution

Question 7.
(i) (x² + 1)\(\frac { dy }{ dx }\) + 2xy = 4x²
(ii) (x² – 1)\(\frac { dy }{ dx }\) + 2xy = \(\frac { 2 }{ x² – 1 }\)
Solution:
(i) Given diff. eqn. can be written as,
\(\frac{d y}{d x}+\left(\frac{2 x}{x^2+1}\right) y=\frac{4 x^2}{x^2+1}\)
which is L.D.E in y of first order and is of dv
the form \(\frac { dy }{ dx }\) + Py = Q

which is the required solution.

(ii) Given, (x² – 1)\(\frac { dy }{ dx }\) + 2xy = \(\frac{2}{x^2-1}\)

which is the required solution.

Question 8.
(i) x logx\(\frac { dy }{ dx }\) + y = 2logx
(ii) x log x\(\frac { dy }{ dx }\) + y = \(\frac { 2 }{ x }\) log x
(iii) x\(\frac { dy }{ dx }\) + 2y = x² log x
Solution:
(i) Given x logx\(\frac { dy }{ dx }\) + y = 2logx

which is the required solution.

(ii) Given x log x\(\frac { dy }{ dx }\) + y = \(\frac { 2 }{ x }\) log x

which gives the required solution.

(iii) Given x\(\frac { dy }{ dx }\) + 2y = x² log x

which gives the required solution.

Question 9.
x sin \(\frac { dy }{ dx }\) + (xcosx + sinx)y = sin x
Solution:
Given
x sin x \(\frac { dy }{ dx }\) + (x cos x + sin x) y = sin x

which gives the required solution.

Question 10.
\(\frac { dy }{ dx }\) + 2y = xe^4x
Solution:
Given \(\frac { dy }{ dx }\) + 2y = xe^4x
which is L.D.E in y of first order and is of dy
the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = 2 ; Q = xe^4x

which is the required solution.

Question 11.
sin x \(\frac { dy }{ dx }\) + 2y + sin x (1 + cos x) = 0
Solution:
Given
sin x \(\frac { dy }{ dx }\) + 2y + sin x (1 + cos x) = 0

which gives the required solution.

Question 12.
(i) (1 + x²)\(\frac { dy }{ dx }\) + y = \(e^{\tan ^{-1} x}\)
(ii) (1 + x²)\(\frac { dy }{ dx }\) + y = tan^-1x
(iii) \(\frac{d y}{d x}+\frac{y}{x}=\cos x+\frac{\sin x}{x}\)
(iv) y’ + y = \(\frac{1+x \log x}{x}\)
Solution:
(i) Given (1 + x²)\(\frac { dy }{ dx }\) + y = \(e^{\tan ^{-1} x}\)
⇒ \(\frac{d y}{d x}+\left(\frac{1}{1+x^2}\right) y=\frac{e^{\tan ^{-1} x}}{1+x^2}\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q

(ii) Given (1 + x²)\(\frac { dy }{ dx }\) + y = tan^-1x
⇒ \(\frac{d y}{d x}+\left(\frac{1}{1+x^2}\right) y=\frac{\tan ^{-1} x}{1+x^2}\)
which is linear diff. eqn. in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.

which gives the required solution.

(iii) Given differential eqn. be
\(\frac{d y}{d x}+\frac{y}{x}=\cos x+\frac{\sin x}{x}\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q

which is the required solution.

(iv) Given, y’ + y = \(\frac{1+x \log x}{x}\)
which is L.D.E in y of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q

gives the required solution.

Question 13.
y log y \(\frac { dx }{ dy }\) + x – logy = 0
Solution:
Given y log y \(\frac { dx }{ dy }\) + x – logy = 0

which is the required solution.

Question 14.
\(\frac { dx }{ dy }\) + y sec x = tan x
Solution:
Given \(\frac { dx }{ dy }\) + y secx = tan x
which is L.D.E in y of first order and is of the form \(\frac { dx }{ dy }\) + Py = Q.
Here P = sec x ; Q = tan x

which gives the required solution.

Question 15.
(x + tan y) dy = sin 2y dx
Solution:
Given diff. eqn. can be written as

which gives the required solution.

Question 16.
Solve the following equations :
(i) \(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) = 2x²
(ii) \(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = sinx, giving the general solution and also the solution for which y = 0 and x = \(\frac { π }{ 2 }\)
Solution:
(i) Given diff. eqn. be, \(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) = 2x²
which is L.D.E in y of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q

(ii) Given \(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = sin x
which is L.D.E in y of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = \(\frac { 1 }{ x }\) ; Q = sin x

which gives the required general solution
given y = 0 when x = \(\frac { π }{ 2 }\)
∴ from (1); we have
0 = – \(\frac { π }{ 2 }\) cos \(\frac { π }{ 2 }\) + sin \(\frac { π }{ 2 }\) + c ⇒ c = – 1
Thus eqn. (1) becomes ;
xy = – x cos x + sin x – 1
which gives the required solution.

Question 17.
(i) \(\frac { dy }{ dx }\) = y tanx – 2sin x
(ii) y’ + y = sin x
(iii) y’ + y = cosx
(iv) y’ + 2y = sin x
(v) 2 \(\frac { dy }{ dx }\) + 4y = sin 2x
Solution:
(i) Given diff. eqn. can be written as
\(\frac { dy }{ dx }\) – y tan x = – 2 sin x
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
Here P = – tan x ; Q = – 2 sin x

which gives the required solution.

(ii) Given, y’ + y = sin x, which is linear in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
Here P = 1 ; Q = sin x

which gives the required solution.

(iii) Given y’ + y = cos x
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = 0
Here P = 1 ; Q = cos x

which gives the required solution.

(iv) Given y’ + 2y = sin x … (1)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = 2 ; Q = sinx

which gives the required solution.

(v) Given diff. eqn. be,
\(\frac { dy }{ dx }\) + 2y = \(\frac { 1 }{ 2 }\) sin 2x … (1)
which is linear in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
Here P = 2 ; Q = \(\frac { 1 }{ 2 }\) sin 2x

which gives the required solution.

Question 18.
(i) (x + 2y²) \(\frac { dy }{ dx }\) = y, y > 0
(ii) (x + 3y²)\(\frac { dy }{ dx }\) = y, given that when x = 2, y = 1.
(iii) (3y² – x)dy = ydx
(iv) y² + (x – \(\frac { 1 }{ y }\))\(\frac { dy }{ dx }\) = 0
(v) (x + y + 1)\(\frac { dy }{ dx }\) = 1
(vi) y dx + (x – y³)dy = 0
Solution:
(i) Given (x + 2y²)\(\frac { dy }{ dx }\) = y, y > 0
⇒ \(\frac { dy }{ dx }\) = \(\frac{x+2 y^2}{y}=\frac{x}{y}+2 y\)
⇒ \(\frac { dx }{ dy }\) – \(\frac { x }{ y }\) = 2y which is linear diff. eqn. in x of first order and is of the form
\(\frac { dx }{ dy }\) + Px = Q.

which gives the required solution.

(ii) Given diff. eqn. be, (x + 3y²)\(\frac { dy }{ dx }\) = y

given that when x = 2, y = 1
∴ from (2); we have 2 = 3 + c ⇒ c = – 1
∴ from (2); we have
x = 3y² – y be the required solution.

(iii) Given (3y² – x)dy = ydx

which gives the required solution.

(iv) Given diff. eqn. be,

which gives the required solution.

(v) Given (x + y + 1)\(\frac { dy }{ dx }\) = 1

which gives the required solution.

(vi) Given diff. eqn. be,

which is the required solution.

Question 19.
Solve the differential equations :
(i) \(\frac { dy }{ dx }\) – 3 y cot x = sin 2x given that y = 2, when x = \(\frac { π }{ 2 }\)
(ii) \(\frac { dy }{ dx }\) + 2y tan x = sin x,
if y = 0 for x = \(\frac { π }{ 3 }\)
(iii) \(\frac { dy }{ dx }\) + x cot y = 2y + y² cot y, (y ≠ 0)
given that x = 0 when y = \(\frac { π }{ 2 }\)
(iv) x\(\frac { dy }{ dx }\) + y = x cos x + sin x, given y(\(\frac { π }{ 2 }\)) = 1
Solution:
(i) Given, \(\frac { dy }{ dx }\) – 3 y cot x = sin 2x … (1)
which is linear in y of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q
Here P = – 3 cot x ; Q = sin 2x

∴ from (1); we have
y = – 2 sin² x + 4 sin³ x
gives the required solution.

(ii) Given \(\frac { dy }{ dx }\) + 2y tan x = sin x
which is linear in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = 2 tan x; Q = sin x

which is the required general solution.
When x = \(\frac { π }{ 3 }\), y = 0 ∴ from (1); we have
0 = \(\frac { 1 }{ 2 }\) + \(\frac { c }{ 4 }\)
⇒ \(\frac { c }{ 4 }\) = – \(\frac { 1 }{ 2 }\)
⇒ c = – 2
Thus eqn. (1) becomes; y = cos x – 2 cos² x
be the required particular solution.

(iii) Given diff. eqn. be dx
\(\frac { dx }{ dy }\) + x cot y = 2y + y² cot y
which is linear is x of first order and is of the form \(\frac { dx }{ dy }\) + Px = Q
Here P = cot y and Q = 2y + y² cot y

be the required solution.

(iv) Given diff. eqn. can be written as
\(\frac { dy }{ dx }\) + \(\frac { 1 }{ x }\)y = cos x + \(\frac { sin x }{ x }\)
which is linear in y of first order and is of the form \(\frac { dx }{ dy }\) + Py = Q.
Here P = \(\frac { 1 }{ x }\) and Q = cos x + \(\frac { sin x }{ x }\)
and solution is given by

∴ eqn, (1) becomes ; y = sin x be the required particular solution.

Question 20.
Solve the differential equation
(1 + x²)\(\frac { dy }{ dx }\) + 2xy – 4x² = 0
subject to the initial condition y (0) = 0.
Solution:
Given diff. eqn. be
(1 + x²)\(\frac { dy }{ dx }\) + 2xy – 4x² = 0
⇒ \(\frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q

Thus eqn. (1) gives ; y (1 + x²) = \(\frac { 4x³ }{ 3 }\)
gives the required particular solution.

Question 21.
Solve \(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\), if when y = 0, x = 1.
Solution:
Given \(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)
which is L.D.E in y of first order and is of
the form \(\frac { dy }{ dx }\)+ Py = Q.

be the required solution.

Question 22.
x\(\frac { dy }{ dx }\) + y = x³
given thaty = 1 when x = 2.
Solution:
Given diflf. eqn. can be written as
\(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = x²
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = \(\frac { 1 }{ x }\) and Q = x²

be the required particular solution.

Question 23.
Find the particular solution of differential equation \(\frac{d y}{d x}=-\frac{x+y \cos x}{1+\sin x}\), given that y = 1, when x = 0.
Solution:
Given \(\frac{d y}{d x}=-\frac{x}{1+\sin x}-\frac{y \cos x}{1+\sin x}\)

Given y = 1 when x = 0
∴ from (1); we have
1 (1 + 0) = 0 + c ⇒ c = 1
∴ eqn. (1) becomes ;
y (1 + sin x) = – \(\frac { x² }{ 2 }\) + 1
be the required solution.