Question

Question: In Problems 17-22, solve the initial value problem.

$\mathbf{\left(}\mathbf{cosx}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{ysinx}\mathbf{=}\mathbf{2}{\mathbf{xcos}}^{\mathbf{2}}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{15}\sqrt{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{32}}$

Short answer

The initial value of the given equation is $\mathrm{y}={\mathrm{x}}^2\mathrm{cosx}-{\mathrm{\pi }}^2\mathrm{cosx}$.

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,

$\left(\mathrm{cosx}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ysinx}=2{\mathrm{xcos}}^2\mathrm{x},\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{15\sqrt{2}{\mathrm{\pi }}^2}{32}$

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

$$\begin{gathered} \left(\mathrm{cosx}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ysinx}=2{\mathrm{xcos}}^2\mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\frac{\mathrm{sinx}}{\mathrm{cosx}}=2\mathrm{x}\frac{{\cos }^2\mathrm{x}}{\mathrm{cosx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytanx}=2\mathrm{xcosx}...........\left(2\right) \end{gathered}$$

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

Where $\mathrm{P}(\mathrm{x})=tan\mathrm{x}$.

Then,

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

Simplification method

Multiply in equation (2)

$\begin{array}{rcl}\mathrm{sexc}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ysecxtanx}& =& 2\mathrm{xcosxsecx}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{secxy}\right]& =& 2\mathrm{x}\end{array}$

Integrating both sides.

role="math" localid="1664167783573" $\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{secxy}\right]& =& \int 2\mathrm{xdx}\\ \mathrm{ysecx}& =& \frac{2{\mathrm{x}}^2}{2}+\mathrm{C}\\ & =& {\mathrm{x}}^2+\mathrm{C}\\ \mathrm{y}& =& \frac{1}{\mathrm{secx}}{\mathrm{x}}^2+\frac{\mathrm{C}}{\mathrm{secx}}\\ \mathrm{y}& =& {\mathrm{x}}^2\mathrm{cosx}+\mathrm{cosxC}........\left(3\right)\end{array}$

Find the initial value

Given that, .

$\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{15\sqrt{2}{\mathrm{\pi }}^2}{32}$

Then, $\mathrm{x}=\frac{\mathrm{\pi }}{4}$and $\mathrm{y}=-\frac{15\sqrt{2}{\mathrm{\pi }}^2}{32}$.

Substitute the value in equation (3) to get the value of C.

$\begin{array}{rcl}\mathrm{y}& =& {\mathrm{x}}^2\mathrm{cosx}+\mathrm{cosxC}\\ -\frac{15\sqrt{2}{\mathrm{\pi }}^2}{32}& =& {\left(\frac{\mathrm{\pi }}{4}\right)}^2\cos \left(\frac{\mathrm{\pi }}{4}\right)+\mathrm{Ccos}\left(\frac{\mathrm{\pi }}{4}\right)\\ & =& \frac{{\mathrm{\pi }}^2}{16}\left(\frac{\sqrt{2}}{2}\right)+\mathrm{C}\left(\frac{\sqrt{2}}{2}\right)\\ -\frac{30{\mathrm{\pi }}^2}{32}& =& \frac{{\mathrm{\pi }}^2}{16}+\mathrm{C}\\ \frac{-30{\mathrm{\pi }}^2-2{\mathrm{\pi }}^2}{32}& =& \mathrm{C}\\ -\frac{32{\mathrm{\pi }}^2}{32}& =& \mathrm{C}\\ \mathrm{C}& =& {\mathrm{\pi }}^2\end{array}$

Substitute the value of C in equation (2).

$\mathrm{y}={\mathrm{x}}^2\mathrm{cosx}-{\mathrm{\pi }}^2\mathrm{cosx}$

So, the solution is$\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{cosx}\mathbf{-}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{cosx}$