Let [ x= (x^ y^)] be the solution of the differential equation [2 x^2 y ~d y-(1-x y^2) d x…

Question

Let $$\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$$ be the solution of the differential equation $$2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$$. Then $$\alpha+\beta-\gamma$$ equals :

Accepted Answer

$\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\ & \Rightarrow 2 x^2 y \frac{d y}{d x}-1+x y^2=0 \\ & \Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x^2}+\frac{y^2}{x}=0 \\ & \Rightarrow 2 y \frac{d y}{d x}+\frac{y^2}{x}=\frac{1}{x^2} \text { (L.D.E) } \\ & Let, y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \\ & \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \\ & \text { I.F }=e^{\int \frac{1}{x} d x}=x\end{aligned}$

So, the general solution is

$$
\begin{aligned}
& \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\
& \Rightarrow y^2 \cdot x=\ln x+c \\
& \text { Also,y(2) }=\sqrt{\log _e 2} \\
& \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\
& \Rightarrow y^2 x=\ln x+\ln 2 \\
& \Rightarrow y^2 x=\ln 2 x \\
& 2 x=\exp \left(x^1 y^2\right)
\end{aligned}
$$

By compare it with given solution we get,

$$
\begin{aligned}
& \alpha=2,\beta=1, \gamma=2 \\
& \Rightarrow \alpha+\beta-\gamma=2+1-2=1
\end{aligned}
$$

Solutions · 1