Question

y^-2dy/dx + y^-1= 2x

Answer 1

The given DE can be written in the form:

$$\frac{1}{y^2}\dfrac{dy}{dx} + \frac{1}{y}= 2x$$

which is a Bernoulli type of first order DE.

$$\text{Solve Bernoulli's equation }\frac{1}{y(x)}+\frac{\frac{d y(x)}{d x}}{y(x)^{2}}=2 x:\\[3mm] \text{Multiply both sides by }-1 :\\ -\frac{\frac{d y(x)}{d x}}{y(x)^{2}}-\frac{1}{y(x)}=-2 x\\ \text{Let }v(x)=\frac{1}{y(x)}, \text{ which gives }\frac{d v(x)}{d x}=-\frac{\frac{d y(x)}{d x}}{y(x)^{2}}:\\ \frac{d v(x)}{d x}-v(x)=-2 x \text{ Let }\mu(x)=e^{\int-1 d x}=e^{-x}.\\[3mm] \text{Multiply both sides by }\mu(x) :\\ e^{-x} \frac{d v(x)}{d x}-e^{-x} v(x)=-2 e^{-x} x\\ \text{Substitute }-e^{-x}=\frac{d}{d x}\left(e^{-x}\right):\\ e^{-x} \frac{d v(x)}{d x}+\frac{d}{d x}\left(e^{-x}\right) v(x)=-2 e^{-x} x\\$$

$$\text{Apply the reverse product rule } f \frac{d g}{d x}+g \frac{d f}{d x}=\frac{d}{d x}(f g)\text{ to the left-hand side}:\\ \frac{d}{d x}\left(e^{-x} v(x)\right)=-2 e^{-x} x\\[3mm] \text{Integrate both sides with respect to }x :\\[3mm] \int \frac{d}{d x}\left(e^{-x} v(x)\right) d x=\int-2 e^{-x} x d x\\[3mm] \text{Evaluate the integrals}:\\ e^{-x} v(x)=-2 e^{-x}(-x-1)+c_{1},\text{ where } c_{1} \text{ is an arbitrary constant.}\\[3mm] \text{Divide both sides by }\mu(x)=e^{-x}:\\[3mm] v(x)=2 x+c_{1} e^{x}+2$$

Solving for $y(x)$, we have:

$$y(x)=\frac{1}{v(x)}=\frac{1}{2 x+c_{1} e^{x}+2}$$

Which is the required solution of the given DE.