Solve yp=frac{e^-x_{1}}{(D+1)^3}x^2

$$y p = \frac { e ^ { - x _ { 1 } } } { ( D + 1 ) ^ { 3 } } x ^ { 2 }$$

Solve for p

$$\left\{\begin{matrix}p=\frac{x^{2}}{y\left(D+1\right)^{3}e^{x_{1}}}\text{, }&D\neq -1\text{ and }y\neq 0\\p\in \mathrm{R}\text{, }&x=0\text{ and }y=0\text{ and }D\neq -1\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(D+1\right)^{3}.

$$yp\left(D+1\right)^{3}=e^{-x_{1}}x^{2}$$

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(D+1\right)^{3}.

$$yp\left(D^{3}+3D^{2}+3D+1\right)=e^{-x_{1}}x^{2}$$

Use the distributive property to multiply yp by D^{3}+3D^{2}+3D+1.

$$ypD^{3}+3ypD^{2}+3ypD+yp=e^{-x_{1}}x^{2}$$

Combine all terms containing p.

$$\left(yD^{3}+3yD^{2}+3yD+y\right)p=e^{-x_{1}}x^{2}$$

The equation is in standard form.

$$\left(yD^{3}+3yD^{2}+3Dy+y\right)p=\frac{x^{2}}{e^{x_{1}}}$$

Divide both sides by yD^{3}+3yD^{2}+3yD+y.

$$\frac{\left(yD^{3}+3yD^{2}+3Dy+y\right)p}{yD^{3}+3yD^{2}+3Dy+y}=\frac{x^{2}}{e^{x_{1}}\left(yD^{3}+3yD^{2}+3Dy+y\right)}$$

Dividing by yD^{3}+3yD^{2}+3yD+y undoes the multiplication by yD^{3}+3yD^{2}+3yD+y.

$$p=\frac{x^{2}}{e^{x_{1}}\left(yD^{3}+3yD^{2}+3Dy+y\right)}$$

Divide \frac{x^{2}}{e^{x_{1}}} by yD^{3}+3yD^{2}+3yD+y.

$$p=\frac{x^{2}}{y\left(D+1\right)^{3}e^{x_{1}}}$$

Solve for D

$$\left\{\begin{matrix}D=\frac{x^{\frac{2}{3}}e^{-\frac{x_{1}}{3}}}{\sqrt[3]{py}}-1\text{, }&x\neq 0\text{ and }p\neq 0\text{ and }y\neq 0\\D\neq -1\text{, }&\left(y=0\text{ or }p=0\right)\text{ and }x=0\end{matrix}\right.$$