Question

In Problems 21-26 solve the given initial-value problem.

$\left(y^{2}cosx-3x^{2}y-2x\right)\mathit{d}\mathit{x}\mathbf{+}\left(2ysinx-x^3+lny\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathit{e}$

Short answer

The initial value of the problem is$y^{2}sinx-x^{3}y-x^2+ylny-y=0$

Step-by-step solution

Given Information

$\begin{array}{l}Thegivenequationis,\left(y^2\cos x-3x^{2}y-2x\right)dx+\left(2y\sin x-x^3+\ln y\right)dy=0\\ Comparethegivenequationwith,Mdx+Ndy=0\\ M(x,y)=y^{2}cosx-3x^{2}y\\ N(x,y)=2ysinx-x^3+lny\end{array}$

Condition for exactness

$AnequationoftheformMdx+Ndy=0issaidtobeexact,if\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial t}$ .

$$\begin{gathered} \frac{\partial M}{\partial y}=2ycosx-3x^2 \\ \frac{\partial N}{\partial x}=2ycosx-3x^2 \end{gathered}$$

Since, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$.

Therefore, the equation is exact

Find the solution

Now, consider the equation as, $f(x,y)=y^{2}sinx-x^{3}y-x^2+g\left(y\right)$

Now integrate $M(x,y)=\frac{\partial f}{\partial x}=y^{2}cosx-3x^{2}y-2x$

$\frac{\partial f}{\partial y}=2ysinx-x^3+g^{¢}\left(y\right)$

Take

$\begin{array}{l}\frac{\partial f}{\partial y}\\ g^{¢}\left(y\right)=lny\end{array}$

$\begin{array}{l}g\left(y\right)=ylny-y\\ g^{¢}\left(y\right)=lny\end{array}$

So, the solution is,

$y^{2}sinx-x^{3}y-x^2+ylny-y=c$

Substitute initial condition$y\left(0\right)=e$

$\begin{array}{c}0-0-0+e-e=c\\ c=0\end{array}$

Hence, the required solution is,$y^{2}sinx-x^{3}y-x^2+ylny-y=0$

$y\left(0\right)=e$

$\begin{array}{c}0-0-0+e-e=c\\ c=0\end{array}$