Get comprehensive solutions for NCERT Class 12 Maths Exercise 9.5. Learn to solve first-order Linear Differential Equations using the Integrating Factor (IF) method. Practice finding the IF, calculating general solutions, and determining particular solutions by applying initial conditions. Includes solving equations linear in $x$ and the Bernoulli’s form. Essential for complete mastery of Chapter 9.
Table of Contents
This exercise focuses on solving First Order Linear Differential Equations of the form:
$$\mathbf{\frac{dy}{dx} + P(x)y = Q(x)}$$
The solution involves two main steps:
- Find the Integrating Factor (IF): $$\mathbf{IF = e^{\int P(x) dx}}$$
- Find the General Solution: $$\mathbf{y \cdot (IF) = \int Q(x) \cdot (IF) dx + C}$$
Part 1: General Solutions (1-12)
1. $\frac{dy}{dx} + 2y = \sin x$
This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 2$ and $Q(x) = \sin x$.
- IF: $IF = e^{\int 2 dx} = \mathbf{e^{2x}}$
- General Solution: $y \cdot e^{2x} = \int \sin x \cdot e^{2x} dx + C$
To solve $I = \int e^{2x} \sin x dx$, use Integration by Parts twice (or the cyclic integral formula):
$I = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) – b \cos(bx))$. Here $a=2, b=1$.
$$I = \frac{e^{2x}}{4 + 1} (2 \sin x – 1 \cos x) = \frac{e^{2x}}{5} (2 \sin x – \cos x)$$
$$y e^{2x} = \frac{e^{2x}}{5} (2 \sin x – \cos x) + C$$
Divide by $e^{2x}$:
$$\mathbf{y = \frac{1}{5} (2 \sin x – \cos x) + C e^{-2x}}$$
2. $\frac{dy}{dx} + 3y = e^{-2x}$
$P(x) = 3$, $Q(x) = e^{-2x}$.
- IF: $IF = e^{\int 3 dx} = \mathbf{e^{3x}}$
- General Solution: $y \cdot e^{3x} = \int e^{-2x} \cdot e^{3x} dx + C$$$y e^{3x} = \int e^{x} dx + C$$$$y e^{3x} = e^x + C$$$$\mathbf{y = e^{-2x} + C e^{-3x}}$$
3. $\frac{dy}{dx} + \frac{y}{x} = x^2$
$P(x) = \frac{1}{x}$, $Q(x) = x^2$.
- IF: $IF = e^{\int (1/x) dx} = e^{\log|x|} = \mathbf{x}$ (assuming $x>0$)
- General Solution: $y \cdot x = \int x^2 \cdot x dx + C$$$xy = \int x^3 dx + C$$$$xy = \frac{x^4}{4} + C$$$$\mathbf{y = \frac{x^3}{4} + \frac{C}{x}}$$
4. $\frac{dy}{dx} + (\sec x) y = \tan x$
$P(x) = \sec x$, $Q(x) = \tan x$.
- IF: $IF = e^{\int \sec x dx} = \mathbf{e^{\log|\sec x + \tan x|} = \sec x + \tan x}$
- General Solution: $y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C$$$y (\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C$$$$y (\sec x + \tan x) = \int (\sec x \tan x + (\sec^2 x – 1)) dx + C$$$$y (\sec x + \tan x) = \sec x + \tan x – x + C$$$$\mathbf{y = 1 – \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}}$$
5. $\cos^2 x \frac{dy}{dx} + y = \tan x$
Divide by $\cos^2 x$: $\frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\tan x}{\cos^2 x}$
$$\frac{dy}{dx} + (\sec^2 x) y = \tan x \sec^2 x$$
$P(x) = \sec^2 x$, $Q(x) = \tan x \sec^2 x$.
- IF: $IF = e^{\int \sec^2 x dx} = \mathbf{e^{\tan x}}$
- General Solution: $y \cdot e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} dx + C$Use substitution: $t = \tan x$, $dt = \sec^2 x dx$.$$y e^{\tan x} = \int t e^t dt + C$$Use Integration by Parts on $\int t e^t dt$: $t e^t – \int e^t dt = t e^t – e^t = e^t (t – 1)$.$$y e^{\tan x} = e^{\tan x} (\tan x – 1) + C$$$$\mathbf{y = \tan x – 1 + C e^{-\tan x}}$$
6. $x \frac{dy}{dx} + 2y = x^2 \log x$
Divide by $x$: $\frac{dy}{dx} + \frac{2}{x} y = x \log x$
$P(x) = \frac{2}{x}$, $Q(x) = x \log x$.
- IF: $IF = e^{\int (2/x) dx} = e^{2 \log|x|} = e^{\log(x^2)} = \mathbf{x^2}$
- General Solution: $y \cdot x^2 = \int (x \log x) x^2 dx + C$$$y x^2 = \int x^3 \log x dx + C$$Use Integration by Parts: $u = \log x$, $dv = x^3 dx$. $du = \frac{1}{x} dx$, $v = \frac{x^4}{4}$.$$y x^2 = \log x \left(\frac{x^4}{4}\right) – \int \frac{x^4}{4} \cdot \frac{1}{x} dx + C$$$$y x^2 = \frac{x^4}{4} \log x – \frac{1}{4} \int x^3 dx + C$$$$y x^2 = \frac{x^4}{4} \log x – \frac{1}{4} \frac{x^4}{4} + C$$$$y x^2 = \frac{x^4}{4} \log x – \frac{x^4}{16} + C$$$$\mathbf{y = \frac{x^2}{4} \log x – \frac{x^2}{16} + \frac{C}{x^2}}$$
7. $x \log x \frac{dy}{dx} + y = 2 \log x$
Divide by $x \log x$: $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2 \log x}{x \log x} = \frac{2}{x}$
$P(x) = \frac{1}{x \log x}$, $Q(x) = \frac{2}{x}$.
- IF: $IF = e^{\int \frac{1}{x \log x} dx}$. Let $t = \log x$, $dt = \frac{1}{x} dx$. $\int \frac{dt}{t} = \log|t|$.$$IF = e^{\log|\log x|} = \mathbf{\log x}$$
- General Solution: $y \cdot \log x = \int \frac{2}{x} \cdot (\log x) dx + C$Use substitution: $t = \log x$, $dt = \frac{1}{x} dx$.$$y \log x = 2 \int t dt + C$$$$y \log x = 2 \frac{t^2}{2} + C$$$$y \log x = (\log x)^2 + C$$$$\mathbf{y = \log x + \frac{C}{\log x}}$$
8. $(1 + x^2) dy + 2xy dx = \cot x dx$
Divide by $(1 + x^2) dx$: $\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{\cot x}{1 + x^2}$
$$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2}$$
$P(x) = \frac{2x}{1 + x^2}$, $Q(x) = \frac{\cot x}{1 + x^2}$.
- IF: $IF = e^{\int \frac{2x}{1 + x^2} dx}$. $\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$.$$IF = e^{\log|1 + x^2|} = \mathbf{1 + x^2}$$
- General Solution: $y \cdot (1 + x^2) = \int \frac{\cot x}{1 + x^2} \cdot (1 + x^2) dx + C$$$y (1 + x^2) = \int \cot x dx + C$$$$y (1 + x^2) = \log|\sin x| + C$$$$\mathbf{y = \frac{\log|\sin x| + C}{1 + x^2}}$$
9. $x \frac{dy}{dx} + y – x + xy \cot x = 0$
$x \frac{dy}{dx} + y(1 + x \cot x) = x$
Divide by $x$: $\frac{dy}{dx} + \frac{1 + x \cot x}{x} y = 1$
$$\frac{dy}{dx} + \left( \frac{1}{x} + \cot x \right) y = 1$$
$P(x) = \frac{1}{x} + \cot x$, $Q(x) = 1$.
- IF: $IF = e^{\int (\frac{1}{x} + \cot x) dx} = e^{\log|x| + \log|\sin x|} = e^{\log|x \sin x|} = \mathbf{x \sin x}$
- General Solution: $y (x \sin x) = \int 1 \cdot (x \sin x) dx + C$Use Integration by Parts on $\int x \sin x dx$: $u=x, dv=\sin x dx$.$$\int x \sin x dx = -x \cos x – \int (-\cos x) dx = -x \cos x + \sin x$$$$y x \sin x = -x \cos x + \sin x + C$$$$\mathbf{y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}}$$
10. $(x + y) \frac{dy}{dx} = 1$
Rewrite the DE w.r.t $y$: $\frac{dx}{dy} = x + y$
$$\mathbf{\frac{dx}{dy} – x = y}$$
This is linear in $x$: $\frac{dx}{dy} + P(y)x = Q(y)$, with $P(y) = -1$ and $Q(y) = y$.
- IF: $IF = e^{\int -1 dy} = \mathbf{e^{-y}}$
- General Solution: $x \cdot e^{-y} = \int y \cdot e^{-y} dy + C$Use Integration by Parts on $\int y e^{-y} dy$: $u=y, dv=e^{-y} dy$.$$\int y e^{-y} dy = -y e^{-y} – \int (-e^{-y}) dy = -y e^{-y} – e^{-y} = -e^{-y} (y + 1)$$$$x e^{-y} = -e^{-y} (y + 1) + C$$Multiply by $e^y$:$$\mathbf{x = -(y + 1) + C e^y \quad \text{or} \quad x + y + 1 = C e^y}$$
11. $y dx + (x – y^2) dy = 0$
This cannot be put into $\frac{dy}{dx} + P(x)y = Q(x)$. Rewrite as $\frac{dx}{dy}$:
$$y \frac{dx}{dy} + x – y^2 = 0$$
Divide by $y$: $\frac{dx}{dy} + \frac{1}{y} x = y$
This is linear in $x$: $P(y) = \frac{1}{y}$, $Q(y) = y$.
- IF: $IF = e^{\int (1/y) dy} = e^{\log|y|} = \mathbf{y}$
- General Solution: $x \cdot y = \int y \cdot y dy + C$$$xy = \int y^2 dy + C$$$$xy = \frac{y^3}{3} + C$$$$\mathbf{x = \frac{y^2}{3} + \frac{C}{y}}$$
12. $\frac{dy}{dx} + 2x y = y^2 e^{x^2}$
This is a Bernoulli’s equation. Divide by $y^2$:
$$y^{-2} \frac{dy}{dx} + 2x y^{-1} = e^{x^2}$$
Substitute $z = y^{-1}$. Differentiate $z$ w.r.t $x$: $\frac{dz}{dx} = -1 y^{-2} \frac{dy}{dx}$.
So, $y^{-2} \frac{dy}{dx} = -\frac{dz}{dx}$.
$$-\frac{dz}{dx} + 2x z = e^{x^2}$$
Multiply by $-1$:
$$\frac{dz}{dx} – 2x z = -e^{x^2}$$
This is linear in $z$: $P(x) = -2x$, $Q(x) = -e^{x^2}$.
- IF: $IF = e^{\int -2x dx} = e^{-x^2}$
- General Solution: $z \cdot e^{-x^2} = \int (-e^{x^2}) \cdot e^{-x^2} dx + C$$$z e^{-x^2} = \int -1 dx + C$$$$z e^{-x^2} = -x + C$$Substitute back $z = 1/y$:$$\frac{1}{y} e^{-x^2} = -x + C$$$$\mathbf{\frac{1}{y} = (C – x) e^{x^2}}$$
Part 2: Particular Solutions (13-15)
13. $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$
$P(x) = 2 \tan x$, $Q(x) = \sin x$.
- IF: $IF = e^{\int 2 \tan x dx} = e^{2 \log|\sec x|} = e^{\log|\sec^2 x|} = \mathbf{\sec^2 x}$
- General Solution: $y \sec^2 x = \int \sin x \sec^2 x dx + C$$$y \sec^2 x = \int \frac{\sin x}{\cos^2 x} dx = \int \tan x \sec x dx + C$$$$y \sec^2 x = \sec x + C$$
- Apply Initial Condition: $y = 0$ when $x = \frac{\pi}{3}$.$$0 \cdot \sec^2\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + C$$$$\sec(\frac{\pi}{3}) = 2$$$$0 = 2 + C \implies C = -2$$$$y \sec^2 x = \sec x – 2$$$$\mathbf{y = \frac{1}{\sec x} – \frac{2}{\sec^2 x} \quad \text{or} \quad y = \cos x – 2 \cos^2 x}$$
14. $(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$; $y = 0$ when $x = 1$
Divide by $(1 + x^2)$: $\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2}$
$P(x) = \frac{2x}{1 + x^2}$, $Q(x) = \frac{1}{(1 + x^2)^2}$.
- IF: $IF = e^{\int \frac{2x}{1 + x^2} dx} = e^{\log(1 + x^2)} = \mathbf{1 + x^2}$
- General Solution: $y \cdot (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) dx + C$$$y (1 + x^2) = \int \frac{1}{1 + x^2} dx + C$$$$y (1 + x^2) = \tan^{-1} x + C$$
- Apply Initial Condition: $y = 0$ when $x = 1$.$$0 (1 + 1^2) = \tan^{-1}(1) + C$$$$0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}$$$$\mathbf{y (1 + x^2) = \tan^{-1} x – \frac{\pi}{4}}$$
15. $\frac{dy}{dx} – 3y \cot x = \sin 2x$; $y = 2$ when $x = \frac{\pi}{2}$
$P(x) = -3 \cot x$, $Q(x) = \sin 2x$.
- IF: $IF = e^{\int -3 \cot x dx} = e^{-3 \log|\sin x|} = e^{\log|\sin^{-3} x|} = \mathbf{\csc^3 x}$
- General Solution: $y \csc^3 x = \int \sin 2x \cdot \csc^3 x dx + C$$$\sin 2x \csc^3 x = (2 \sin x \cos x) \frac{1}{\sin^3 x} = 2 \frac{\cos x}{\sin^2 x} = 2 \cot x \csc x$$$$y \csc^3 x = \int 2 \cot x \csc x dx + C$$$$y \csc^3 x = -2 \csc x + C$$
- Apply Initial Condition: $y = 2$ when $x = \frac{\pi}{2}$.$$\csc(\frac{\pi}{2}) = 1$$$$2 \cdot 1^3 = -2 (1) + C$$$$2 = -2 + C \implies C = 4$$$$y \csc^3 x = -2 \csc x + 4$$$$\mathbf{y = -2 \frac{\csc x}{\csc^3 x} + \frac{4}{\csc^3 x} \quad \text{or} \quad y = -2 \sin^2 x + 4 \sin^3 x}$$
Part 3: Applications and Problems (16-17)
16. Slope of tangent at $(x, y)$ is equal to the sum of the coordinates. Curve passes through $(0, 0)$.
The condition is $\frac{dy}{dx} = x + y$.
$$\mathbf{\frac{dy}{dx} – y = x}$$
$P(x) = -1$, $Q(x) = x$.
- IF: $IF = e^{\int -1 dx} = e^{-x}$
- General Solution: $y e^{-x} = \int x e^{-x} dx + C$Use Integration by Parts on $\int x e^{-x} dx$: $-x e^{-x} – \int (-e^{-x}) dx = -x e^{-x} – e^{-x} = -e^{-x} (x + 1)$.$$y e^{-x} = -e^{-x} (x + 1) + C$$$$y = -(x + 1) + C e^x$$$$y + x + 1 = C e^x$$
- Apply Initial Condition: Curve passes through $(0, 0)$.$$0 + 0 + 1 = C e^0 \implies 1 = C$$$$\mathbf{y + x + 1 = e^x \quad \text{or} \quad y = e^x – x – 1}$$
17. Sum of coordinates exceeds the magnitude of the slope of the tangent by 5. Curve passes through $(0, 2)$.
The condition is: Sum of coordinates $-$ Magnitude of slope $= 5$.
$$(x + y) – \left|\frac{dy}{dx}\right| = 5$$
Since the slope is not explicitly given a sign, we consider the most common case for these problems where the relation is linear: $\frac{dy}{dx} = x + y – 5$.
$$\mathbf{\frac{dy}{dx} – y = x – 5}$$
$P(x) = -1$, $Q(x) = x – 5$.
- IF: $IF = e^{\int -1 dx} = e^{-x}$
- General Solution: $y e^{-x} = \int (x – 5) e^{-x} dx + C$Using $\int (ax+b) e^{-x} dx = -e^{-x} (ax + a + b)$:$\int (x – 5) e^{-x} dx = -e^{-x} (x – 5 + 1) = -e^{-x} (x – 4)$$$y e^{-x} = -e^{-x} (x – 4) + C$$$$y = -(x – 4) + C e^x$$$$y + x – 4 = C e^x$$
- Apply Initial Condition: Curve passes through $(0, 2)$.$$2 + 0 – 4 = C e^0 \implies -2 = C$$$$\mathbf{y + x – 4 = -2 e^x \quad \text{or} \quad y = 4 – x – 2 e^x}$$
Part 4: Multiple Choice Questions (18-19)
18. The Integrating Factor (IF) of $\frac{dy}{dx} – \frac{2}{x^2} y = x^2$ is
The equation given in the exercise image is actually $2x \frac{dy}{dx} – y = x^2$ (a typo in the printed version). Assuming the correct DE from the text is $\frac{dy}{dx} – \frac{2}{x} y = x^2$ (a very common problem):
- $2x \frac{dy}{dx} – y = x^2$. Divide by $2x$: $\frac{dy}{dx} – \frac{1}{2x} y = \frac{x}{2}$. $P(x) = -\frac{1}{2x}$.$$IF = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \log x} = e^{\log x^{-1/2}} = \frac{1}{\sqrt{x}}$$ (Not an option)
Assuming the DE is $x \frac{dy}{dx} – 2y = x^2$:
- $\frac{dy}{dx} – \frac{2}{x} y = x$. $P(x) = -\frac{2}{x}$.$$IF = e^{\int -\frac{2}{x} dx} = e^{-2 \log x} = e^{\log x^{-2}} = \mathbf{x^{-2} = \frac{1}{x^2}}$$ (Not an option)
Assuming the DE is $\frac{dy}{dx} – \frac{2}{x} y = x^2$ as implied by the simplest rearrangement of a similar problem:
- $P(x) = -\frac{2}{x}$.$$IF = e^{\int -\frac{2}{x} dx} = e^{-2 \log x} = e^{\log x^{-2}} = \frac{1}{x^2}$$ (Not an option)
Revisiting the original problem $2x \frac{dy}{dx} – y = x^2$ and the options: Since (C) $\frac{1}{x}$ is an option, let’s see which problem gives this IF.
If $P(x) = -\frac{1}{x}$, then $IF = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \mathbf{\frac{1}{x}}$.
This IF corresponds to $\frac{dy}{dx} – \frac{1}{x} y = Q(x)$, which is obtained by dividing $x \frac{dy}{dx} – y = Q(x)$ by $x$. The question likely had a coefficient of $x$ for $\frac{dy}{dx}$.
Given the likely typo in the question and the available options:
If the DE was $x \frac{dy}{dx} – y = Q(x)$, the IF is $\frac{1}{x}$. We select the IF that is an option and is commonly derived from a slight modification of the provided DE.
The most likely intended answer, given the choices, is (C) $\frac{1}{x}$.
19. The Integrating Factor of $\frac{dx}{dy} + \frac{1}{1 – y^2} x = \frac{y}{1 – y^2}$ is
The DE is linear in $x$: $\frac{dx}{dy} + P(y) x = Q(y)$.
$$P(y) = \frac{y}{1 – y^2}$$
(Typo in Q: should be $\frac{y}{1 – y^2} x$).
Assuming $P(y) = \frac{-y}{1 – y^2}$ from the provided text: $\frac{dx}{dy} – \frac{y}{1 – y^2} x = \dots$
Let’s assume the text meant $P(y) = \frac{y}{1 – y^2}$ (as written in the formula part, $y x$ term):
$$IF = e^{\int \frac{y}{1 – y^2} dy}$$
Let $t = 1 – y^2$, $dt = -2y dy$. $\frac{y dy}{1 – y^2} = -\frac{1}{2} \frac{dt}{t}$.
$$\int \frac{y}{1 – y^2} dy = -\frac{1}{2} \log|1 – y^2| = \log|(1 – y^2)^{-1/2}|$$
$$IF = e^{\log|(1 – y^2)^{-1/2}|} = (1 – y^2)^{-1/2} = \mathbf{\frac{1}{\sqrt{1 – y^2}}}$$
The correct answer is (D) $\frac{1}{\sqrt{1 – y^2}}$.