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In each of the following differential equation show that it is homogeneous and solve it. \frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0
CBSE | Class 12 | exercise 20 | Homogeneous Differential Equations | Maths | RS Aggarwal
In each of the following differential equation show that it is homogeneous and solve it. \frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0

Solution:

\begin{array}{l} \Rightarrow \frac{d y}{d x}=-\frac{x-2 y}{2 x-y}=-\frac{1-2^{\frac{z}{x}}{x}}{2-\frac{z}{x}} \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}
\Rightarrow the given differential eq. is a homogenous eq.
Solution of the given differential eq. is :
\begin{array}{l} \text { Put } y=v x \\ \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow v+x \frac{d v}{d x}=-\frac{1-2 \frac{v x}{x}}{2-\frac{v x}{x}} \\ \Rightarrow v+x \frac{d v}{d x}=-\frac{1-2 v}{2-v} \\ \Rightarrow x \frac{d v}{d x}=-\frac{1-2 v}{2-v}-v=-\frac{1-2 v+2 v-v^{2}}{2-v}=-\frac{1-v^{2}}{2-v} \\ \Rightarrow \frac{2-v}{v^{2}-1} d v=\frac{d x}{x} \\ \Rightarrow \frac{v-2}{v^{2}-1} d v=-\frac{d x}{x} \\ \Rightarrow \frac{v}{v^{2}-1} d v-\frac{2}{v^{2}-1} d v=-\frac{d x}{x} \end{array}
On integrating both sides we obtain:
\begin{array}{l} \Rightarrow \int \frac{v}{v^{2}-1} d v-\int \frac{2}{v^{2}-1} d v=-\int \frac{d x}{x}+c \\ \Rightarrow \frac{\ln \left|v^{2}-1\right|}{2}-2 \times \frac{1}{2} \ln \left|\frac{v-1}{v+1}\right|=-\ln |x|+\ln |c| \end{array}
By resubstituting the value of y=v x we have
\begin{array}{l} \Rightarrow \frac{\ln \left|\left(\frac{y}{x}\right)^{2}-1\right|}{2}-2 \times \frac{1}{2} \ln \left|\frac{\left(\frac{y}{x}\right)-1}{\left(\frac{y}{x}\right)+1}\right|=-\ln |x|+\ln |c| \\ \Rightarrow(y-x)=C(y+x)^{3} \end{array}

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