Question

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

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

Short answer

The general solution to the given equation is $\mathrm{y}=2{\mathrm{x}}^2+\mathrm{xln}\left|\mathrm{x}\right|+\mathrm{Cx}$.

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 $\mathbf{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$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 $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$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,

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

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

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

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

Where $P\left(x\right)=-\frac{1}{x}$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}-\frac{1}{x}\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\ln \mathrm{x}}\\ & =& \frac{1}{\mathrm{x}}\end{array}$

Simplification method

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

$$\begin{gathered} \frac{1}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\frac{\mathrm{y}}{\mathrm{x}}=\frac{1}{\mathrm{x}}\left(2\mathrm{x}+1\right) \\ \frac{1}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{{\mathrm{x}}^2}=2+\frac{1}{\mathrm{x}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{\mathrm{x}}\mathrm{y}\right]=2+\frac{1}{\mathrm{x}} \end{gathered}$$

Integrating both sides.

$\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{\mathrm{x}}\mathrm{y}\right]& =& {\int }_{}^{}\left(2+\frac{1}{\mathrm{x}}\right)\mathrm{dx}\\ \frac{\mathrm{y}}{\mathrm{x}}& =& 2\mathrm{x}+\ln \left|\mathrm{x}\right|+\mathrm{C}\end{array}$

So, the solution is $\begin{array}{rcl}\frac{\mathbf{y}}{\mathbf{x}}& \mathbf{=}& \mathbf{2}\mathbf{x}\mathbf{+}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}\end{array}$.