Question

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

$\mathbf{y}\frac{\mathbf{dx}}{\mathbf{dy}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{5}{\mathbf{y}}^{\mathbf{3}}$

Short answer

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

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 $\mathbf{\mu }\left(x\right)$ by 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.

 Step 2: Solve the given equation

Given that,

$\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}+2\mathrm{x}=5{\mathrm{y}}^3.......\left(1\right)$

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

role="math" localid="1664261627869" $$\begin{gathered} \mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}+2\mathrm{x}=5{\mathrm{y}}^3 \\ \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{2\mathrm{x}}{\mathrm{y}}=5{\mathrm{y}}^2.......\left(2\right) \end{gathered}$$

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

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

Then,

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

 Step 3: Simplification method

Multiply $\mathrm{\mu }\left(\mathrm{t}\right)$in equation (2)

role="math" localid="1664262043390" $\begin{array}{rcl}{\mathrm{y}}^2\frac{\mathrm{dx}}{\mathrm{dy}}+{\mathrm{y}}^2\frac{2\mathrm{x}}{\mathrm{y}}& =& 5{\mathrm{y}}^2{\mathrm{y}}^2\\ {\mathrm{y}}^2\frac{\mathrm{dx}}{\mathrm{dy}}+2\mathrm{xy}& =& 5{\mathrm{y}}^4\\ \frac{\mathrm{d}}{\mathrm{dy}}\left[{\mathrm{y}}^2\mathrm{x}\right]& =& 5{\mathrm{y}}^4\end{array}$

Integrating both sides.

$\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dy}}\left[{\mathrm{y}}^2\mathrm{x}\right]dx& =& {\int }_{}^{}5{\mathrm{y}}^4\mathrm{dx}\\ {\mathrm{y}}^2\mathrm{x}& =& \frac{5{\mathrm{y}}^5}{5}+\mathrm{C}\\ \mathrm{x}& =& {\mathrm{y}}^3+{\mathrm{Cy}}^{-2}\\ & & \end{array}$

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