Question

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

$\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{3}\left(y+x^2\right)\mathbf{=}\frac{\mathbf{sinx}}{\mathbf{x}}$

Short answer

The general solution to the given equation is $\mathrm{y}=-\frac{\mathrm{cosx}}{{\mathrm{x}}^2}+\frac{\mathrm{sinx}}{{\mathrm{x}}^3}-\frac{3}{5}{\mathrm{x}}^2+\frac{\mathrm{C}}{{\mathrm{x}}^3}$.

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$\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{exp}\mathbf{\left[}{\mathbf{\int }}_{}^{}\mathbf{P}\left(x\right)\mathbf{dx}\mathbf{\right]}$.
• 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,

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

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

localid="1664263451058" $\begin{array}{rcl}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+3(\mathrm{y}+{\mathrm{x}}^2)& =& \frac{\mathrm{sinx}}{\mathrm{x}}\\ \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{y}+3{\mathrm{x}}^2& =& \frac{\mathrm{sinx}}{\mathrm{x}}\\ \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{y}& =& \frac{\mathrm{sinx}}{\mathrm{x}}-3{\mathrm{x}}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}+3\frac{\mathrm{y}}{\mathrm{x}}& =& \frac{\mathrm{sinx}}{{\mathrm{x}}^2}-3\mathrm{x}.........\left(2\right)\end{array}$

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

Where $P\left(x\right)=\frac{3}{x}$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}\frac{3}{\mathrm{x}}\mathrm{dx}}\\ & =& {\mathrm{e}}^{3\ln \left|\mathrm{x}\right|}\\ & =& {\mathrm{x}}^3\end{array}$

 Step 3: Simplification method

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

$\begin{array}{rcl}{x}^3\frac{\mathrm{dy}}{\mathrm{dx}}+3x^3\frac{\mathrm{y}}{\mathrm{x}}& =& x^3\frac{\mathrm{sinx}}{{\mathrm{x}}^2}-3x^3\\ {\mathrm{x}}^3\frac{\mathrm{dy}}{\mathrm{dx}}+3{\mathrm{x}}^2\mathrm{y}& =& {\mathrm{x}}^2\mathrm{sinx}-3{\mathrm{x}}^4\end{array}$

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^3\mathrm{y}\right]& =& {\mathrm{x}}^2\mathrm{sinx}-3{\mathrm{x}}^4\end{array}$

Integrating both sides.

$\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^3\mathrm{y}\right]dx& =& \int \left(\mathrm{xsinx}-3{\mathrm{x}}^4\right)\mathrm{dx}\\ {\mathrm{x}}^3\mathrm{y}& =& {\int }_{}^{}\mathrm{xsinxdx}-\int 3{\mathrm{x}}^4\mathrm{dx}\\ {\mathrm{x}}^3\mathrm{y}& =& {\int }_{}^{}\mathrm{xsinxdx}-\frac{3{\mathrm{x}}^5}{5}+\mathrm{C}......\left(3\right)\end{array}$

Find the value of ${\int }_{}^{}\mathrm{xsinxdx}$separately.

Let

$$\begin{gathered} \mathrm{u}=\mathrm{x},\mathrm{dv}=\mathrm{xsinxdx} \\ \mathrm{du}=\mathrm{dx},\mathrm{v}=-\mathrm{cosx} \end{gathered}$$

$\begin{array}{rcl}{\int }_{}^{}\mathrm{xsinxdx}& =& -\mathrm{xcosx}+{\int }_{}^{}\mathrm{cosx}\\ & =& -\mathrm{xcosx}+\mathrm{sinx}\end{array}$

Substitute the value in equation (3).

$\begin{array}{rcl}{\mathrm{x}}^3\mathrm{y}& =& -x\cos x+\sin x-\frac{3{\mathrm{x}}^5}{5}+\mathrm{C}\\ \mathrm{y}& =& -\frac{\mathrm{cosx}}{{\mathrm{x}}^2}+\frac{\mathrm{sinx}}{{\mathrm{x}}^3}-\frac{3{\mathrm{x}}^2}{5}+\frac{\mathrm{C}}{{\mathrm{x}}^3}\end{array}$

So, the solution is $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{-}\frac{\mathbf{cosx}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{sinx}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{C}}{{\mathbf{x}}^{\mathbf{3}}}\end{array}$.