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On putting $\frac{y}{x}=v$ the differential equation $\frac{d y}{d x}=\frac{2 x y-y^2}{2 x y-x^2}$ is transformed to |
$x(2 v-1) d x=3 v(v-1) d x$ $x(2 v-1) d v=3 v(1-v) d x$ $x(1-2 v) d v=\left(v^2-2 v\right) d x$ $x(1-2 v) d v=\left(v^2-2 v\right) d x$ |
$x(2 v-1) d v=3 v(1-v) d x$ |
Putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$, given differential equation reduces to $v+x \frac{d v}{d x}=\frac{2 v-v^2}{2 v-1} \Rightarrow 3 v(1-v) d x=x(2 v-1) d v$ |
