Question

Question: In Problems 7-16, obtain the general solution to the equation.

$\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}$.

Short answer

The general solution to the given equation is$\mathrm{y}=-{\mathrm{x}}^{-3}+{\mathrm{Cx}}^{-2}$.

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathit{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by$\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$. Here C is an arbitrary constant.

Solve the given equation

Given that,

$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}={\mathrm{x}}^{-3}...........\left(1\right)$

Write the equation (1) into standard form of linear equation.

$$\begin{gathered} \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}={\mathrm{x}}^{-3} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+2\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^{-4}........\left(2\right) \end{gathered}$$

Calculate the integrating factor of $\mu \left(x\right)$.

Where $P\left(x\right)=\frac{2}{x}$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}\frac{2}{\mathrm{x}}\mathrm{dx}}\\ & =& {\mathrm{e}}^{2\ln \left|x\right|}\\ & =& {\mathrm{x}}^2\end{array}$

Simplification method

Multiply $\mu \left(x\right)$in equation (2)

$\begin{array}{rcl}{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}+2{\mathrm{x}}^2\frac{\mathrm{y}}{\mathrm{x}}& =& {\mathrm{x}}^2{\mathrm{x}}^{-4}\\ {\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}& =& {\mathrm{x}}^{-2}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^2\mathrm{y}\right]& =& {\mathrm{x}}^{-2}\\ & & \end{array}$

Integrating both sides.

$\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^2\mathrm{y}\right]dx& =& {\int }_{}^{}{\mathrm{x}}^{-2}\mathrm{dx}\\ {\mathrm{x}}^2\mathrm{y}& =& {\mathrm{x}}^{-1}+\mathrm{C}\\ \mathrm{y}& =& {\mathrm{x}}^{-3}+{\mathrm{Cx}}^2\end{array}$

So, the solution is $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& {\mathbf{x}}^{\mathbf{-}\mathbf{3}}\mathbf{+}{\mathbf{Cx}}^{\mathbf{2}}\end{array}$.