Question

Determine whether the given differential equation is exact. If it is exact, solve it.

$\left(x-y^3+y^{2}sinx\right)\mathit{d}\mathit{x}\mathbf{=}\mathbf{(}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{y}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{)}\mathit{d}\mathit{y}$

Short answer

The solution is$-y^{3}x-y^{2}cos\left(x\right)+\frac{1}{2}{x}^2=c$

Step-by-step solution

Given information

The given equation is$\left(x-y^3+y^{2}sinx\right)dx=(3x{y}^2+2ycosx)dy$. An equation of the form$Mdx+Ndy=f\left(\text{x}\right)$ is exact only if it satisfies$\frac{\partial \text{M}}{\partial \text{x}}=\frac{\partial \text{N}}{\partial \text{y}}$

Determining the exactness of the differential equation

Find M and N using given equation.

$\begin{array}{l}M(x,y)=x-y^3+y^2\mathrm{Sin}\left(x\right)\\ N(x,y)=-3x{y}^2-2y\mathrm{Cos}\left(x\right)\end{array}$

Find role="math" localid="1663831036094" $\frac{\partial M}{\partial y}$and $\frac{\partial N}{\partial X}$

$\begin{array}{l}\frac{\partial M}{\partial y}=-3y^2+2y\mathrm{Sin}\left(x\right)\\ \frac{\partial N}{\partial x}=-3y^2+2y\mathrm{Sin}\left(x\right)\end{array}$

As , $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$so, the equation is exact

Integrate M and N

The solution of the differential equation is a function such that $\frac{\partial f}{\partial x}=\text{M}$and $\frac{\partial f}{\partial y}=\text{N}$. Here $\frac{\partial f}{\partial x}=x-y^3+y^2\mathrm{Sin}\left(x\right)$and $\frac{\partial f}{\partial y}=-3x{y}^2-2y\mathrm{Cos}\left(x\right)$. After integrating the first of these equations, we get$f\left(x,y\right)=g\left(y\right)+\frac{x^2}{2}$

$N(x,y)$is obtained by taking the partial derivative of the previous expression with respect to y and setting the result to $N(x,y)$

$\begin{array}{c}\frac{\partial f}{\partial y}=g\text{'}\left(y\right)\\ =-3x{y}^2-2y\mathrm{Cos}\left(x\right)\end{array}$

On comparing these two equations we get

$\begin{array}{c}g\text{'}\left(y\right)=-3x{y}^2-2y\mathrm{Cos}\left(x\right)\\ g\left(y\right)=-x{y}^3-y^2\mathrm{Cos}\left(x\right)\end{array}$

Therefore, function $f\left(x,y\right)$becomes $f\left(x,y\right)=-y^{3}x-y^{2}Cos\left(x\right)+\frac{1}{2}{x}^2$, As a result, the implicit solution of the problem is$-y^{3}x-y^{2}Cos\left(x\right)+\frac{1}{2}{x}^2=C$