Question

dy/dx = 5y + (e ^−2x)(y^−2)

Answer 1

Solution:

$\begin{aligned} &\frac{d y}{d x}-5 y=e^{-2 x} y^{-2} \\ &\Longrightarrow y^{2} \frac{d y}{d x}-5 y^{3}=e^{-2 x} \\ &\Longrightarrow 3 y^{2} \frac{d y}{d x}-15 y^{3}=3 e^{-2 x} \\ &\text { Let, } \mathrm{z}=\mathrm{y}^{\wedge} 3 \\ &\frac{d z}{d x}=3 y^{2} \\ &\Longrightarrow \frac{d z}{d x}-15 z=3 e^{-2 x} \ldots(i) \end{aligned}$

Now this differential equation is of the form

$\frac{d y}{d x}+P(x) y=Q(x)$

here, $P(x)=-15$

$Q(x)=3 e^{-2 x}$

So,

$\text { Integrating Factor }=e^{\int P(x) d x}=e^{\int-15 d x}=e^{-15 x}$

Multiplying Integrating factor to the both sides of equation (i),

 $e^{-15 x} \frac{d z}{d x}-15 e^{-15 x} z=3 e^{-17 x}$

$\begin{aligned} &\Longrightarrow \frac{d\left(e^{-15 x} z\right)}{d x}=3 e^{-17 x} \\ &\Longrightarrow d\left(e^{-15 x} z\right)=3 e^{-17 x} d x \end{aligned}$

Integrating both sides,

$\begin{aligned} &\Longrightarrow \int d\left(e^{-15 x} z\right)=\int 3 e^{-17 x} d x \\ &\Longrightarrow e^{-15 x} z=-\frac{3}{17} e^{-17 x}+C \\ &\Longrightarrow y^{3} e^{-15 x}=-\frac{3}{17} e^{-17 x}+C \end{aligned}$