91Ó°ÊÓ

• 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 $\mathbf{μ}\left(x\right)$by the formula .
• Multiply the equation in standard form by $\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{\pm Êμ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{μ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

Given that,

$\left(\mathrm{sinx}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+y\cos x=\mathrm{xsinx},\mathrm{y}\left(\frac{\mathrm{Ï€}}{2}\right)=2........\left(1\right)$

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

$$\begin{gathered} \left(\mathrm{sinx}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycosx}=\mathrm{xsinx} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\frac{\mathrm{cosx}}{\mathrm{sinx}}=\mathrm{x}\frac{\mathrm{sinx}}{\mathrm{sinx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycotx}=\mathrm{x}...........\left(2\right) \end{gathered}$$

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

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

Then,

$\begin{array}{rcl}\mathrm{μ}\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{∫}_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{∫}_{}^{}\mathrm{cotxdx}}\\ & =& {\mathrm{e}}^{\left(\ln \left|\sin x\right|\right)}\\ & =& \sin x\end{array}$

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

$$\begin{gathered} \sin x\frac{\mathrm{dy}}{\mathrm{dx}}+y\sin x\mathrm{cotx}=\mathrm{xsinx} \\ \mathrm{sinx}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycosx}=\mathrm{xsinx} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{ysinx}\right]=\mathrm{xsinx} \end{gathered}$$

Integrating both sides.

${∫}_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{ysinx}\right]\mathrm{dx}={∫}_{}^{}\mathrm{xsinxdx}........\left(3\right)$

Find the value ofrole="math" localid="1664169607457" ${∫}_{}^{}\mathrm{xsinxdx}$separately.

Let $\mathrm{u}=\mathrm{x},\mathrm{dv}=\mathrm{sinxdx}$

$\mathrm{du}=\mathrm{dx},\mathrm{v}=-\mathrm{cosx}$

Then,

$\begin{array}{rcl}{∫}_{}^{}\mathrm{xsinxdx}& =& -\mathrm{xcosx}+∫\mathrm{cosxdx}\\ & =& -\mathrm{xcosx}+\mathrm{sinx}+\mathrm{C}\end{array}$

Substitute the value in equation (3)

role="math" localid="1664170021261" $\begin{array}{rcl}{∫}_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{ysinx}\right]\mathrm{dx}& =& ∫\mathrm{xsinxdx}\\ \mathrm{ysinx}& =& -\mathrm{xcosx}+\mathrm{sinx}+\mathrm{C}\\ \mathrm{y}& =& -\mathrm{x}\frac{\mathrm{cosx}}{\mathrm{sinx}}+\frac{\sin }{\mathrm{sinx}}\mathrm{x}+\frac{\mathrm{C}}{\mathrm{sinx}}\\ \mathrm{y}& =& -\mathrm{xcotx}+1+\frac{\mathrm{C}}{\mathrm{sinx}}..........\left(4\right)\end{array}$

Given that, $y\left(\frac{\mathrm{Ï€}}{2}\right)=2$.

Then, $x=\frac{\mathrm{Ï€}}{2}$and $y=2$.

Substitute the value in equation (4) to get the value of C.

$\begin{array}{rcl}\mathrm{y}& =& -\mathrm{xcotx}+1+\frac{\mathrm{C}}{\mathrm{sinx}}\\ 2& =& -\frac{\mathrm{Ï€}}{2}\cot \left(\frac{\mathrm{Ï€}}{2}\right)+1+\frac{\mathrm{C}}{\sin \left(\frac{\mathrm{Ï€}}{2}\right)}\\ 2& =& 0+1+\mathrm{C}\\ \mathrm{C}& =& 1\end{array}$

Substitute the value of C in equation (4).

So, the solution is$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{-}\mathbf{xcotx}\mathbf{+}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{sinx}}\end{array}$