Question

In Problems 37 and 38 solve the given initial-value problem by finding, as in Example 4, an appropriate integrating factor.

37.$\mathit{x}\mathit{d}\mathit{x}\mathbf{+}\left(x^{2}y+4y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is $e^{y^2}\left(x^2+4\right)=20$.

Step-by-step solution

Determine whether the given equation is exact or not

The given equation is $xdx+\left(x^{2}y+4y\right)dy=0$.

Identify M and N.

$\begin{array}{l}M=x\\ N=x^{2}y+4y\end{array}$

Find $M_y$and $N_x$.

$\begin{array}{c}{M}_y=0\\ N_x=2xy\end{array}$

Note that so this is not yet an exact differential equation.

Find $\frac{N_x-M_y}{M}$and $\frac{N_y-M_x}{M}$.

$$\begin{gathered} \begin{array}{c}\frac{M_y-N_x}{M}=\frac{0-2xy}{x^{2}y+4y}\\ =\frac{-2x}{x^2+4}\end{array} \\ \begin{array}{c}\frac{N_x-M_y}{M}=\frac{2xy-0}{x}\\ =\frac{2xy}{x}\\ =2y\end{array} \end{gathered}$$

Find the integration factor

Now, find the integrating factor as follow:

$\begin{array}{rcl}\mu \left(y\right)& =& {\mathrm{e}}^{{\int }_{}\frac{{\displaystyle N_x-M_y}}{{\displaystyle M}}\mathrm{dy}}\\ & =& {\mathrm{e}}^{2\mathrm{ydy}}\\ & =& {\mathrm{e}}^{{\mathrm{y}}^2}\end{array}$

Multiply M and N by the integration factor

$\begin{array}{c}\mu \left(y\right)M+\mu \left(y\right)N=0\\ x{e}^{y^2}dx+\left(x^{2}y{e}^{y^2}+4y{e}^{y^2}\right)dy=0\end{array}$

Identify M and N again and take the partial derivatives.

$\begin{array}{c}M=x{e}^{y^2}\\ N=x^{2}y{e}^{y^2}+4y{e}^{y^2}\end{array}$

The partial derivatives equal one another so this is now an exact differential equation.

$$\begin{gathered} M_y=2xy{e}^{y^2} \\ {N}_x=2xy{e}^{y^2} \end{gathered}$$

Here we take the integral of M to get our function. Note the integral can also be taken of N in which case you would integrate with respect to y.

$\begin{array}{c}f(x,y)=\int x{e}^{y^2}dx\\ =\frac{1}{2}{x}^2{e}^{y^2}+g\left(y\right)\\ \frac{\partial f}{\partial y}=x^{2}y{e}^{y^2}+g\text{'}\left(y\right)\end{array}$

Where $g\text{'}\left(y\right)=4y{e}^{y^2}$.

Take the integral of $g\text{'}\left(y\right)$to $g\left(y\right)$get Integrate with u substitution.

$$\begin{gathered} g\left(y\right)=\int 4y{e}^{y^2}dy \\ Letu=y^2\Rightarrow du=2ydy\Rightarrow 2du=4ydy \end{gathered}$$

Then, we have,

$\begin{array}{c}g\left(y\right)=2\int e^{u}du\\ =2e^u\\ =2e^{y^2}\end{array}$

Then, the solution becomes $\frac{1}{2}{x}^2{e}^{y^2}+2e^{y^2}=C$.

Now, find the value if constant by using the given initial condition $y\left(4\right)=0$.

$\begin{array}{c}\frac{1}{2}(4{)}^2{e}^{{\left(0\right)}^2}+2e^{{\left(0\right)}^2}=C\\ \frac{1}{2}\left(16\right)\left(1\right)+2\left(1\right)=C\\ 8+2=C\\ 10=C\end{array}$

Plug C back into the equation. Note that these three equations are equivalent.

$\begin{array}{c}\frac{1}{2}{x}^2{e}^{y^2}+2e^{y^2}=10\\ \iff x^2{e}^{y^2}+4e^{y^2}=20\\ \iff e^{y^2}\left(x^2+4\right)=20\end{array}$

Hence, the final required solution is$e^{y^2}\left(x^2+4\right)=20$.