In problem 17 –26, solve the initial value problem.

$x^2\frac{dy}{dx}=\frac{4x^2-x-2}{(x+1)(y+1)},y\left(1\right)=1$

A first-order ordinary differential equation having the form $\frac{dy}{dx}=f(x,y),y\left(x_0\right)=y_0$is referred to as an initial value problem. The solution of an initial value problem is a particular curve that satisfies the given differential equation and passes through the point $\left(x_0{y}_0\right)$ . The particular solution is determined from the general solution. By plugging the initial condition into the general condition, the numerical value of the integrating constant is determined. It results in a particular curve belonging to the family of curves.

The given equation is

$x^2\frac{dy}{dx}=\frac{4x^2-x-2}{(x+1)(y+1)}............\left(1\right)$

After separating the variables, equation (1) can be written as

(y+1)dy=$\frac{4x^2-x-2}{(x+1)(y+1)}dx...........\left(2\right)$

In order to proceed further, let us apply the partial fraction decomposition to simplify the function role="math" localid="1664357475525" $\frac{4x^2-x-2}{x^2(x+1)}$.

Now,

$\frac{4x^2-x-2}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$

Here, A, B, C are the constants we are going to determine. Now,

$$\begin{gathered} \frac{4x^2-x-2}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ \frac{4x^2-x-2}{x^2(x+1)}=\frac{x^{2}C+x(x+1)A+(x+1)B}{x^2(x+1)} \\ 4x^2-x-2=x^{2}C+x(x+1)A+(x+1)B \\ 4x^2-x-2=x^{2}A+x^{2}C+xA+xB+B \\ 4x^2-x-2=x(A+C)+x(A+B)+B \\ Comparethecoefficientsof{x}^2,xandthecons\tan tterms.Itgivesvaluesof \\ A+C=4,A+B=-1,B=-2 \\ A=1,B=-2,C=3 \\ \frac{4x^2-x-2}{x^2(x+1)}=\frac{1}{x}+(-\frac{2}{x^2})+\frac{3}{x+1} \\ Integratebothsidesofequation.Itresults, \\ \frac{1}{2}{y}^2+y=\ln \left|x{(x+1)}^3\right|+\frac{2}{x}+C \end{gathered}$$

Now, apply the initial condition by plugging in the general solution. It follows,

$$\begin{gathered} \frac{1}{2}+1=2+\ln 8+C \\ C=1-\ln 8+\frac{1}{2} \\ C=\frac{3}{2}-\ln 8 \end{gathered}$$

Therefore the solution of the given initial value problem is

$\frac{1}{2}{y}^2+y=ln\left|x{(x+1)}^3\right|+\frac{2}{x}+\frac{3}{2}-ln8$ .