Exercise 9.4 Practice
Chapter 9: Differential Equations – NCERT Solutions (Rationalised)
Topic: Homogeneous Differential Equations
Q1
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Show that the differential equation $(x^2 + xy) dy = (x^2 + y^2) dx$ is homogeneous and solve it.
$\frac{dy}{dx} = \frac{x^2+y^2}{x^2+xy}$. Let $F(x,y) = \frac{x^2+y^2}{x^2+xy}$. Since $F(\lambda x, \lambda y) = F(x,y)$, it is homogeneous.
Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.
$v + x\frac{dv}{dx} = \frac{x^2+v^2x^2}{x^2+vx^2} = \frac{1+v^2}{1+v}$.
$x\frac{dv}{dx} = \frac{1+v^2}{1+v} - v = \frac{1-v}{1+v}$.
$\int \frac{1+v}{1-v} dv = \int \frac{dx}{x} \Rightarrow \int (-1 + \frac{2}{1-v}) dv = \log|x| + C$.
$-v - 2\log|1-v| = \log|x| + C$. Substitute $v=y/x$.
$\boxed{(x-y)^2 = Cxe^{-y/x}}$
Q2
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Solve: $y' = \frac{x+y}{x}$
Put $y = vx \Rightarrow v + x\frac{dv}{dx} = \frac{x+vx}{x} = 1+v$.
$x\frac{dv}{dx} = 1 \Rightarrow dv = \frac{dx}{x}$.
$v = \log|x| + C$.
$\boxed{y = x\log|x| + Cx}$
Q3
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Solve: $(x-y) dy - (x+y) dx = 0$
$\frac{dy}{dx} = \frac{x+y}{x-y}$. Put $y=vx$.
$v + x\frac{dv}{dx} = \frac{1+v}{1-v} \Rightarrow x\frac{dv}{dx} = \frac{1+v^2}{1-v}$.
$\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x} \Rightarrow \tan^{-1}v - \frac{1}{2}\log(1+v^2) = \log|x| + C$.
$\tan^{-1}(y/x) - \frac{1}{2}\log(\frac{x^2+y^2}{x^2}) = \log|x| + C$.
$\boxed{\tan^{-1}(\frac{y}{x}) = \frac{1}{2}\log(x^2+y^2) + C}$
Q4
00:00
Solve: $(x^2 - y^2) dx + 2xy dy = 0$
$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. Put $y=vx$.
$v + x\frac{dv}{dx} = \frac{v^2-1}{2v} \Rightarrow x\frac{dv}{dx} = -\frac{1+v^2}{2v}$.
$\int \frac{2v}{1+v^2} dv = -\int \frac{dx}{x}$.
$\log(1+v^2) = -\log|x| + \log C \Rightarrow 1+\frac{y^2}{x^2} = \frac{C}{x}$.
$\boxed{x^2 + y^2 = Cx}$
Q5
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Solve: $x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy$
$\frac{dy}{dx} = 1 - 2(\frac{y}{x})^2 + \frac{y}{x}$. Put $y=vx$.
$v + x\frac{dv}{dx} = 1 - 2v^2 + v \Rightarrow x\frac{dv}{dx} = 1 - 2v^2$.
$\int \frac{dv}{1-(\sqrt{2}v)^2} = \int \frac{dx}{x}$.
$\frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}| = \log|x| + C$.
$\boxed{\frac{1}{2\sqrt{2}} \log|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}| = \log|x| + C}$
Q6
00:00
Solve: $x dy - y dx = \sqrt{x^2 + y^2} dx$
$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1+(\frac{y}{x})^2}$. Put $y=vx$.
$v + x\frac{dv}{dx} = v + \sqrt{1+v^2} \Rightarrow \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x}$.
$\log|v+\sqrt{1+v^2}| = \log|x| + \log C$.
$\frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx$.
$\boxed{y + \sqrt{x^2+y^2} = Cx^2}$
Q7
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Solve: $\{x \cos(\frac{y}{x}) + y \sin(\frac{y}{x})\} y dx = \{y \sin(\frac{y}{x}) - x \cos(\frac{y}{x})\} x dy$
$\frac{dy}{dx} = \frac{y}{x} \cdot \frac{x \cos(y/x) + y \sin(y/x)}{y \sin(y/x) - x \cos(y/x)}$. Put $y=vx$.
Simplifying leads to $\frac{v \sin v - \cos v}{v \cos v} dv = \frac{2dx}{x}$.
Integrate: $\log|\sec v| - \log|v| = 2\log|x| + \log C$.
$\sec(y/x) = Cxy$.
$\boxed{xy \cos(\frac{y}{x}) = K}$
Q8
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Solve: $x \frac{dy}{dx} - y + x \sin(\frac{y}{x}) = 0$
$\frac{dy}{dx} = \frac{y}{x} - \sin(\frac{y}{x})$. Put $y=vx$.
$v + x\frac{dv}{dx} = v - \sin v \Rightarrow \int -\csc v dv = \int \frac{dx}{x}$.
$\log|\csc v + \cot v| = -\log|x| + \log C$.
Alternatively, $\int \frac{dv}{\sin v} = \log|\tan(v/2)|$.
$\boxed{x \tan(\frac{y}{2x}) = C}$
Q9
00:00
Solve: $y dx + x \log(\frac{y}{x}) dy - 2x dy = 0$
Rearrange: $\frac{dx}{dy} = \frac{2x - x\log(y/x)}{y}$. This is homogeneous.
Or treat as $\frac{dy}{dx} = \frac{y}{2x - x\log(y/x)}$. Put $y=vx$.
Leads to $\int \frac{2-\log v}{v(\log v - 1)} dv = \int \frac{dx}{x}$.
Let $u = \log v - 1$. Result: $\log|\log(y/x)-1| - (\log(y/x)-1) = \log|x| + C$.
$\boxed{C y = x e^{\log(y/x)-1} \text{ (Implicit form)}}$
Q10
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Solve: $(1 + e^{x/y}) dx + e^{x/y} (1 - \frac{x}{y}) dy = 0$
Put $x = vy \Rightarrow \frac{dx}{dy} = v + y\frac{dv}{dy}$.
$v + y\frac{dv}{dy} = -\frac{e^v(1-v)}{1+e^v} \Rightarrow y\frac{dv}{dy} = -\frac{v+e^v}{1+e^v}$.
$\int \frac{1+e^v}{v+e^v} dv = -\int \frac{dy}{y}$.
$\log|v+e^v| = -\log|y| + \log C$.
$\boxed{x + y e^{x/y} = C}$
Q11
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Find particular solution: $(x+y) dy + (x-y) dx = 0; y=1$ when $x=1$
$\frac{dy}{dx} = \frac{y-x}{y+x}$. Put $y=vx$.
$\int \frac{v+1}{1+v^2} dv = -\int \frac{dx}{x}$.
$\frac{1}{2}\log(1+v^2) + \tan^{-1}v = -\log|x| + C$.
At $x=1, y=1 \Rightarrow v=1$. $\frac{1}{2}\log 2 + \frac{\pi}{4} = C$.
$\boxed{\tan^{-1}(\frac{y}{x}) + \frac{1}{2}\log(x^2+y^2) = \frac{\pi}{4} + \frac{1}{2}\log 2}$
Q12
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Find particular solution: $x^2 dy + (xy + y^2) dx = 0; y=1$ when $x=1$
$\frac{dy}{dx} = -\frac{xy+y^2}{x^2}$. Put $y=vx$.
$\int \frac{dv}{v(v+2)} = -\int \frac{dx}{x} \Rightarrow \frac{1}{2}\log|\frac{v}{v+2}| = -\log|x| + C$.
At $x=1, y=1 \Rightarrow v=1$. $C = \frac{1}{2}\log(1/3)$.
$\boxed{x^2y = \frac{1}{3}(y+2x)}$
Q13
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Find particular solution: $[x \sin^2(\frac{y}{x}) - y] dx + x dy = 0; y=\frac{\pi}{4}$ when $x=1$
$\frac{dy}{dx} = \frac{y}{x} - \sin^2(\frac{y}{x})$. Put $y=vx$.
$x\frac{dv}{dx} = -\sin^2 v \Rightarrow \int -\csc^2 v dv = \int \frac{dx}{x}$.
$\cot v = \log|x| + C$. At $x=1, y=\pi/4 \Rightarrow C=1$.
$\boxed{\cot(\frac{y}{x}) = \log|x| + 1}$
Q14
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Find particular solution: $\frac{dy}{dx} - \frac{y}{x} + \csc(\frac{y}{x}) = 0; y=0$ when $x=1$
$\frac{dy}{dx} = \frac{y}{x} - \csc(\frac{y}{x})$. Put $y=vx$.
$x\frac{dv}{dx} = -\csc v \Rightarrow \int -\sin v dv = \int \frac{dx}{x}$.
$\cos v = \log|x| + C$. At $x=1, y=0 \Rightarrow C=1$.
$\boxed{\cos(\frac{y}{x}) = \log|x| + 1}$
Q15
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Find particular solution: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0; y=2$ when $x=1$
$\frac{dy}{dx} = \frac{2xy+y^2}{2x^2}$. Put $y=vx$.
$x\frac{dv}{dx} = \frac{v^2}{2} \Rightarrow \int \frac{2}{v^2} dv = \int \frac{dx}{x}$.
$-\frac{2}{v} = \log|x| + C$. At $x=1, y=2 \Rightarrow C=-1$.
$\boxed{y = \frac{2x}{1-\log|x|}}$
Q16
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A homogeneous differential equation of the form $\frac{dx}{dy} = h(\frac{x}{y})$ can be solved by making the substitution:
(A) $y=vx$ (B) $v=yx$ (C) $x=vy$ (D) $x=v$
(A) $y=vx$ (B) $v=yx$ (C) $x=vy$ (D) $x=v$
For $\frac{dx}{dy} = h(\frac{x}{y})$, the dependent variable is $x$. We substitute $x = vy$.
$\boxed{\text{(C) } x=vy}$
Q17
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Which of the following is a homogeneous differential equation?
(A) $(4x+6y+5)dy - (3y+2x+4)dx = 0$
(B) $(xy)dx - (x^3+y^3)dy = 0$
(C) $(x^3+2y^2)dx + 2xy dy = 0$
(D) $y^2 dx + (x^2-xy-y^2) dy = 0$
(A) $(4x+6y+5)dy - (3y+2x+4)dx = 0$
(B) $(xy)dx - (x^3+y^3)dy = 0$
(C) $(x^3+2y^2)dx + 2xy dy = 0$
(D) $y^2 dx + (x^2-xy-y^2) dy = 0$
Check degree of each term.
(A) Constants present, not homogeneous.
(B) $xy$ (deg 2), $x^3$ (deg 3). Not homogeneous.
(C) $x^3$ (deg 3), $y^2$ (deg 2). Not homogeneous.
(D) $y^2$ (deg 2), $x^2$ (deg 2), $xy$ (deg 2), $y^2$ (deg 2). All terms degree 2.
$\boxed{\text{(D)}}$