Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 9 Ex 9.4 page covers all 17 questions, solved step-by-step — confirming a differential equation is homogeneous of degree zero, then applying the substitution y=vx (or x=vy when that's cleaner) to reduce it to a variables-separable equation in v and x.
Questions 1–10 build the substitution technique across a wide range of forms — polynomial ratios, square roots, and equations involving \sin(y/x), \cos(y/x) or e^{x/y}, where picking the right substitution direction (y=vx versus x=vy) makes the integral far more manageable. Questions 11–15 add an initial condition to each equation to pin down the constant and find a particular solution. The exercise closes with two MCQs testing the underlying logic of the method itself.
\dfrac{dy}{dx}=\dfrac{x^2+y^2}{x^2+xy}. Both numerator and denominator are homogeneous of degree 2, so the ratio is homogeneous of degree zero.
Put y=vx, so \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}. Substituting: v+x\dfrac{dv}{dx}=\dfrac{1+v^2}{1+v}.
x\dfrac{dv}{dx}=\dfrac{1+v^2}{1+v}-v=\dfrac{1-v}{1+v}, so \dfrac{1+v}{1-v}dv=\dfrac{dx}{x}.
Integrating (writing \frac{1+v}{1-v}=-1+\frac{2}{1-v}): -v-2\log|1-v|=\log|x|+C. Replacing v=y/x and simplifying:
\dfrac{dy}{dx}=1+\dfrac{y}{x}, a function of y/x alone — homogeneous of degree zero.
Put y=vx: v+x\dfrac{dv}{dx}=1+v \Rightarrow x\dfrac{dv}{dx}=1.
Integrating: v=\log|x|+C. Replacing v=y/x: \dfrac{y}{x}=\log|x|+C.
\dfrac{dy}{dx}=\dfrac{x+y}{x-y}, homogeneous of degree zero. Put y=vx: v+x\dfrac{dv}{dx}=\dfrac{1+v}{1-v}.
x\dfrac{dv}{dx}=\dfrac{1+v^2}{1-v}, so \dfrac{1-v}{1+v^2}dv=\dfrac{dx}{x}.
Integrating: \tan^{-1}v-\dfrac{1}{2}\log(1+v^2)=\log|x|+C. Replacing v=y/x and simplifying the log terms:
\dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy}, homogeneous of degree zero. Put y=vx: v+x\dfrac{dv}{dx}=\dfrac{v^2-1}{2v}.
x\dfrac{dv}{dx}=\dfrac{-(1+v^2)}{2v}, so \dfrac{2v}{1+v^2}dv=-\dfrac{dx}{x}.
Integrating: \log(1+v^2)=-\log|x|+C_1, so x(1+v^2)=C. Replacing v=y/x:
Dividing by x^2: \dfrac{dy}{dx}=1-2\left(\dfrac{y}{x}\right)^2+\dfrac{y}{x}, homogeneous of degree zero.
Put y=vx: v+x\dfrac{dv}{dx}=1-2v^2+v \Rightarrow x\dfrac{dv}{dx}=1-2v^2.
\dfrac{dv}{1-2v^2}=\dfrac{dx}{x}. Integrating using \int\dfrac{dv}{1-2v^2}=\dfrac{1}{2\sqrt2}\log\left|\dfrac{1+\sqrt2v}{1-\sqrt2v}\right| and replacing v=y/x:
\dfrac{dy}{dx}=\dfrac{y+\sqrt{x^2+y^2}}{x}, homogeneous of degree zero. Put y=vx: v+x\dfrac{dv}{dx}=v+\sqrt{1+v^2}.
x\dfrac{dv}{dx}=\sqrt{1+v^2}, so \dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}.
Integrating: \log\left|v+\sqrt{1+v^2}\right|=\log|x|+C_1, so v+\sqrt{1+v^2}=Cx. Replacing v=y/x and clearing the fraction:
Put y=vx throughout. After substitution the equation reduces to v+x\dfrac{dv}{dx}=\dfrac{v(\cos v+v\sin v)}{v\sin v-\cos v}.
Simplifying: x\dfrac{dv}{dx}=\dfrac{2v\cos v}{v\sin v-\cos v}, so \left(\tan v-\dfrac{1}{v}\right)dv=\dfrac{2\,dx}{x}.
Integrating: -\log|\cos v|-\log|v|=2\log|x|+C, i.e. \log|v\cos v|=-2\log|x|+C_1, so v\cos v=\dfrac{C}{x^2}. Replacing v=y/x:
\dfrac{dy}{dx}=\dfrac{y}{x}-\sin\left(\dfrac{y}{x}\right), homogeneous of degree zero. Put y=vx: v+x\dfrac{dv}{dx}=v-\sin v.
x\dfrac{dv}{dx}=-\sin v, so \csc v\,dv=-\dfrac{dx}{x}.
Integrating: \log\left|\tan\dfrac{v}{2}\right|=-\log|x|+C_1, so x\tan\dfrac{v}{2}=C. Replacing v=y/x:
\dfrac{dy}{dx}=\dfrac{y/x}{2-\log(y/x)}, homogeneous of degree zero. Put y=vx: v+x\dfrac{dv}{dx}=\dfrac{v}{2-\log v}.
x\dfrac{dv}{dx}=\dfrac{v(\log v-1)}{2-\log v}, so \dfrac{2-\log v}{v(\log v-1)}dv=\dfrac{dx}{x}.
Substituting t=\log v and integrating: \log|\log v-1|-\log v=\log|x|+C. Combining the logs and simplifying with v=y/x:
Here it's cleaner to put x=vy, so \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}. The equation is homogeneous of degree zero in x/y.
Substituting: v+y\dfrac{dv}{dy}=\dfrac{-e^v(1-v)}{1+e^v}, which simplifies to y\dfrac{dv}{dy}=\dfrac{-(v+e^v)}{1+e^v}.
So \dfrac{1+e^v}{v+e^v}dv=-\dfrac{dy}{y}. Integrating: \log|v+e^v|=-\log|y|+C_1, so y(v+e^v)=C. Replacing v=x/y:
\dfrac{dy}{dx}=\dfrac{y-x}{y+x}, homogeneous. Put y=vx: x\dfrac{dv}{dx}=\dfrac{-(1+v^2)}{1+v}, so \dfrac{1+v}{1+v^2}dv=-\dfrac{dx}{x}.
Integrating: \tan^{-1}v+\dfrac{1}{2}\log(1+v^2)=-\log|x|+C.
At x=1,y=1 (so v=1): \dfrac{\pi}{4}+\dfrac{1}{2}\log 2=C. Replacing v=y/x and simplifying the log terms:
\dfrac{dy}{dx}=-\dfrac{y}{x}-\left(\dfrac{y}{x}\right)^2, homogeneous. Put y=vx: x\dfrac{dv}{dx}=-v(2+v), so \dfrac{dv}{v(v+2)}=-\dfrac{dx}{x}.
By partial fractions and integrating: \dfrac{1}{2}\log\left|\dfrac{v}{v+2}\right|=-\log|x|+C_1, so \dfrac{v}{v+2}=\dfrac{C}{x^2}.
At x=1,y=1 (v=1): \dfrac{1}{3}=C. Replacing v=y/x and simplifying:
\dfrac{dy}{dx}=\dfrac{y}{x}-\sin^2\left(\dfrac{y}{x}\right), homogeneous. Put y=vx: x\dfrac{dv}{dx}=-\sin^2 v, so \csc^2v\,dv=-\dfrac{dx}{x}.
Integrating: -\cot v=-\log|x|+C_1, so \cot v=\log|x|+C.
At x=1,y=\pi/4 (v=\pi/4): \cot(\pi/4)=1=0+C\Rightarrow C=1. Replacing v=y/x:
\dfrac{dy}{dx}=\dfrac{y}{x}-\csc\left(\dfrac{y}{x}\right), homogeneous. Put y=vx: x\dfrac{dv}{dx}=-\csc v, so \sin v\,dv=-\dfrac{dx}{x}.
Integrating: -\cos v=-\log|x|+C_1, so \cos v=\log|x|+C.
At x=1,y=0 (v=0): \cos 0=1=0+C\Rightarrow C=1. Replacing v=y/x:
Rewriting: \dfrac{dy}{dx}=\dfrac{2xy+y^2}{2x^2}=\dfrac{y}{x}+\dfrac{1}{2}\left(\dfrac{y}{x}\right)^2, homogeneous.
Put y=vx: x\dfrac{dv}{dx}=\dfrac{v^2}{2}, so \dfrac{2}{v^2}dv=\dfrac{dx}{x}.
Integrating: -\dfrac{2}{v}=\log|x|+C.
At x=1,y=2 (v=2): -1=0+C\Rightarrow C=-1. Replacing v=y/x: -\dfrac{2x}{y}=\log|x|-1.
When the equation is written as \dfrac{dx}{dy} in terms of x/y, it's y that plays the role of independent variable, so the natural substitution is x=vy.
(A) has stray constants (+5, +4), breaking degree-zero homogeneity. (B) has numerator degree 2 but denominator degree 3, so the ratio isn't degree zero. (C) mixes an x^3 term with a y^2 term — not all terms share the same degree.
(D): every term — y^2, x^2, xy — is degree 2, so F(\lambda x,\lambda y)=\lambda^2F(x,y) throughout; dividing gives a ratio homogeneous of degree zero.
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