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

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.