Regular engagement with ISC Class 12 OP Malhotra Solutions Chapter 17 Differential Equations Ex 17(c) can boost students confidence in the subject.

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

Solve the following differential equations:

Question 1.

\(\frac { dy }{ dx }\) = 5x + 7

Solution:

Question 2.

\(\frac { dy }{ dx }\) = sin x – x

Solution:

Given \(\frac { dy }{ dx }\) = sin x – x

⇒ dy = (sin x – x)dx;

On integrating ; we have

∫ dy = ∫ (sin x – x) dx

⇒ y = – cos x – \(\frac { x² }{ 2 }\) + C,

which is the required solution.

Question 3.

\(\frac { dy }{ dx }\) = x log x

Solution:

Question 4.

\(\frac { dy }{ dx }\) + 2x = e^{3x}

Solution:

Given \(\frac { dy }{ dx }\) + 2x = e^{3x}

⇒ dy = (e^{3x} – 2x) dx ;

On integrating ; we have

∫ dy = ∫ (e^{3x} – 2x) dx

⇒ y = \(\frac{e^{3 x}}{3}\) – x² + C,

which is the required solution.

Question 5.

(x + 1)\(\frac { dy }{ dx }\) = x²

Solution:

Question 6.

(x + 1)²\(\frac { dy }{ dx }\) = xe^{x}

Solution:

Question 7.

\(\frac { dy }{ dx }\) = sin³ x cos² x + xe^{x}

Solution:

Question 8.

\(\frac{d y}{d x}=\frac{1}{\sin ^4 x+\cos ^4 x}\)

Solution:

Question 9.

\(\frac{d y}{d x}=x \sin ^2 x+\frac{1}{x \log x}\)

Solution:

Question 10.

\(\sqrt{a+x}\)dy + xdx = 0

Solution:

Given diff. eqn. be,

which is the required solution.

Question 11.

\(\frac{d y}{d x}=\sqrt{4-y^2}\)

Solution:

Question 12.

\(\frac { dy }{ dx }\) = sec y

Solution:

Given \(\frac { dy }{ dx }\) = sec y ⇒ \(\frac{d y}{d x}=\frac{1}{\cos y}\)

⇒ cos y dy = dx

On integrating ; we have

∫ cos y dy = ∫ dx

⇒ sin y = x + c

be the required solution.

Question 13.

\(\frac{d y}{d x}=2^{-y}\)

Solution:

Given \(\frac{d y}{d x}=2^{-y}\)

⇒ \(\frac{1}{2^{-y}}\) dy = dx

⇒ 2^{y} dy = dx

On integrating, we have

\(\int 2^y d y=\int d x \Rightarrow \frac{2^y}{\log 2}=x+\frac{c}{\log 2}\)

⇒ 2^{y} = x log2 + c

which is the required solution.

Question 14.

Find the particular solution of e^{dy/dx} = x + 1, given that x = 0, y = 3.

Solution:

Give diff. eqn. be \(e \frac{d y}{d x}\) = x + 1

Taking logarithm on both sides, we have

\(\frac { dy }{ dx }\) = log (x + 1) ⇒ dy = log (x + 1) dx

On integrating ; we have

given x = 0, y = 3 ∴ from (1) ; we have

3 = log 1 – 0 + c ⇒ c = 3

Thus eqn. (1) becomes :

y = (x+ 1) log (x + 1) – x + 3

be the required solution.

Question 15.

Find the particular solution of the differential equation log(\(\frac { dy }{ dx }\)) = 3x + 4y, given that y = 0 when x = 0.

Solution:

Given diff. eqn. be,

which is the required solution.