Question

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

38. $\left(x^2+y^2-5\right)\mathit{d}\mathit{x}\mathbf{=}\mathbf{(}\mathit{y}\mathbf{+}\mathit{x}\mathit{y}\mathbf{)}\mathit{d}\mathit{y}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

Short answer

The integrating factor is $\frac{4-y^2}{2{(x+1)}^2}+ln(x+1)+\frac{2}{x+1}=\frac{7}{2}$.

Step-by-step solution

Determine whether the given equation is exact or not

The given equation is, $\left(x^2+y^2-5\right)dx=\left(y+xy\right)dy$.

Identifying M and N from the given equation.

$\begin{array}{c}M=x^2+y^2-5\\ N=-(y+xy)\end{array}$

Finding $M_y$and $N_x$

$$\begin{gathered} M_y=2y \\ {N}_x=-y \end{gathered}$$

Note that, the given equation is not an exact differential equation.

Compute $\frac{M_y-N_x}{N}$.

$\begin{array}{c}\frac{M_y-N_x}{N}=\frac{2y-(-y)}{-(y+xy)}\\ =-\frac{3}{1+x}\end{array}$

Find the integration factor

Since $\frac{M_y-N_x}{N}$only depends on x

Now, find the integrating factor as follow:

$\begin{array}{c}\mu \left(x\right)=e^{-\frac{3}{1\times x}dx}\\ =e^{-3\ln (1+x)}\\ =\frac{1}{{\left(1+x\right)}^3}\end{array}$

Multiply both sides of the differential equation by the integrating factor.

$\frac{1}{{\left(1+x\right)}^3}\left(x^2+y^2-5\right)dx=\frac{1}{{\left(1+x\right)}^3}\left(y+xy\right)dy$

Now, identify M and N.

$$\begin{gathered} M(x,y)=\frac{x^2+y^2-5}{{(x+1)}^3} \\ \begin{array}{c}N(x,y)=-\frac{y+xy}{{(x+1)}^3}\\ =-\frac{y}{{(x+1)}^2}\end{array} \end{gathered}$$

Find the partial derivatives of M and N with respect to y and x respectively.

$\begin{array}{l}\frac{\partial M}{\partial y}=\frac{2y}{{(x+1)}^3}\\ \frac{\partial N}{\partial x}=\frac{2y}{{(x+1)}^3}\end{array}$

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

Find the solution

Now, consider the function $f(x,y)=-\frac{y^2}{2{(x+1)}^2}+g\left(x\right)$.

Integrate $N(x,y)=\frac{\partial f}{\partial y}=-\frac{y}{{(x+1)}^2}$.

role="math" localid="1663836533919" $\frac{\partial f}{\partial x}=\frac{y^2}{{(x+1)}^3}+g^{\varphi }\left(x\right)$

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

$g^{\varphi }\left(x\right)=\frac{x^2-5}{{(x+1)}^3}$, since $M(x,y)=\frac{\partial f}{\partial x},$ we see that

$\begin{array}{c}{g}^{\varphi }\left(x\right)=\frac{x^2}{{(x+1)}^3}-\frac{5}{{(x+1)}^3}\\ =\frac{x^2-5}{{(x+1)}^3}\end{array}$

The, we have $$\begin{gathered} g\left(x\right)=ln(x+1)+\frac{2}{x+1}-\frac{1}{2{(x+1)}^2}-\frac{5}{2{\left(x+1\right)}^2} \\ \begin{array}{c}g\left(x\right)=\ln (x+1)+\frac{2}{x+1}+\frac{2}{{(x+1)}^2}\\ -\frac{y^2}{2{(x+1)}^2}+\ln (x+1)+\frac{2}{x+1}+\frac{2}{{(x+1)}^2}=c\\ \frac{4-y^2}{2{(x+1)}^2}+\ln (x+1)+\frac{2}{x+1}=c\end{array} \end{gathered}$$

Now, find the value of the constant by using the initial condition $y\left(0\right)=1$.

$\begin{array}{c}\frac{4-1}{2}+0+2=c\\ c=\frac{7}{2}\end{array}$

Hence, the required solution is $\frac{4-y^2}{2{(x+1)}^2}+ln(x+1)+\frac{2}{x+1}=\frac{7}{2}$.