Question

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

${\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}}^{\mathbf{2}}\mathit{d}\mathit{x}\mathbf{+}\mathbf{(}\mathbf{2}\mathit{x}\mathit{y}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The answer is$x{y}^2+x^{2}y-y+\frac{1}{3}{x}^3=\frac{4}{3}$

Step-by-step solution

Given Information

The given equation is, ${(x+y)}^{2}dx+(2xy+x^2-1)dy=0,y\left(1\right)=1$.

Compare the given equation with$Mdx+Ndy=0$

$$\begin{gathered} M(x,y)={(x+y)}^2=x^2+2xy+y^2 \\ N(x,y)=2xy+x^2-1 \end{gathered}$$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, 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 $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$.

$$\begin{gathered} \frac{\partial M}{\partial y}=2x+2y \\ \frac{\partial N}{\partial x}=2y+2x \end{gathered}$$

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

Integrate

Now, $f(x,y)=x{y}^2+x^{2}y-y+g\left(x\right)$

Take, $\frac{\partial f}{\partial x}$

$\frac{\partial f}{\partial x}=y^2+2xy+g\text{'}\left(x\right)$

Since then, localid="1668417951032" $M(x,y)=\frac{\partial f}{\partial x}$, We have noticed that localid="1668417954272" $g\text{'}\left(x\right)=x^2$

$\begin{array}{c}\mathrm{g}\text{'}\left(\mathrm{x}\right)={\mathrm{x}}^2\\ \Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{3}{\mathrm{x}}^3\end{array}$

Find the solution

So, the result is$x{y}^2+x^{2}y-y+\frac{1}{3}{x}^3=c$.

It is given that, as an initial condition, so put this value in above equation,

$\begin{array}{c}1+1-1+\frac{1}{3}=c\\ c=\frac{4}{3}\end{array}$

So, the final result is$x{y}^2+x^{2}y-y+\frac{1}{3}{x}^3=\frac{4}{3}$