Question

Solve the given differential equation by undetermined coefficients.

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{3}{\mathbf{y}}^{\text{'}}\mathbf{-}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{x}\left({\mathbf{e}}^{\mathbf{x}}\mathbf{+}\mathbf{1}\right)$

Short answer

$y=C_1{e}^{-5x}+C_2{e}^{2x}-\frac{1}{6}x{e}^x-\frac{5}{36}{e}^x-\frac{1}{10}x-\frac{3}{100}\text{.}$

Step-by-step solution

Definition of homogeneous differential equation

A homogeneous differential equation is an equation containing a differentiation and a function, with a set of variables.

Given data

Consider the differential equation as,

$y^{\text{'}\text{'}}+3y^{\text{'}}-10y=x\left(e^x+1\right)$

The objective is to evaluate the above differential equation by undetermined coefficients. First, calculate the complementary solution as,

Find the complementary solution

The corresponding auxiliary equation is,

$\begin{array}{rcl}{m}^2+3m-10& =& 0\\ m^2+5m-2m-10& =& 0\\ m(m+5)-2(m+5)& =& 0\\ (m+5)(m-2)& =& 0\\ m& =& -5,2\end{array}$

Thus, the complementary solution is, .$y_c=C_1{e}^{-5x}+C_2{e}^{2x}$

Apply differential calculator

The term$x\left(e^x+1\right)$ is annihilated by the differential operator$D^2(D-1{)}^2$ as,

$$\begin{gathered} (D-1{)}^2\left[x{e}^x\right]=0 \\ {D}^2(D-1{)}^2\left[x\left(e^x+1\right)\right]=0 \end{gathered}$$

Apply the differential operator $D^2(D-1{)}^2$to the given differential equation.

Thus,$$\begin{gathered} D^2(D-1{)}^2\left(D^2+3D-10\right)y=D^2(D-1{)}^2\left[x\left(e^x+1\right)\right] \\ {D}^2(D-1{)}^2(D-2)(D+5)y=0 \end{gathered}$$

The auxiliary equation is,

$$\begin{gathered} m^2(m-1{)}^2(m-2)(m+5)=0 \\ m=-5,2,0,0,1,1 \end{gathered}$$

Find particular solution

So, the general solution must be

$y=\underset{y_c}{\underbrace{C_1{e}^{-5x}+C_2{e}^{2x}}}+C_{3}x+C_4+C_{5}x{e}^x+C_6{e}^x$

Thus, the general form of particular solution is,

Let$y_p=Ax{e}^x+B{e}^x+Cx+D$

Then,

$$\begin{gathered} y_p^{\text{'}}=A{e}^x+Ax{e}^x+B{e}^x+C=Ax{e}^x+(A+B)e^x+C \\ {y}_p^{\text{'}\text{'}}=A{e}^x+Ax{e}^x+(A+B)e^x=Ax{e}^x+(2A+B)e^x \end{gathered}$$

Substitution

Substitute the above obtained values in equation (1).

$$\begin{gathered} y_p^{\text{'}\text{'}}+3y_p^{\text{'}}-10y_p=x\left(e^x+1\right) \\ Ax{e}^x+(2A+B)e^x+3\left(Ax{e}^x+(A+B)e^x+C\right) \\ -10\left(Ax{e}^x+B{e}^x+Cx+D\right)=x{e}^x+x \\ -6Ax{e}^x+(5A-6B)e^x+(-10C)x+(3C-10D)=x{e}^x+x \end{gathered}$$

Compare the coefficient

Compare the coefficients of like terms as,

Coefficient of$x{e}^x:-6A=1\Rightarrow A=-\frac{1}{6}$

Coefficient of$e^x:5A-6B=0\Rightarrow 5\left(-\frac{1}{6}\right)-6B=0\Rightarrow B=-\frac{5}{36}$

Coefficient of$x:-10C=1\Rightarrow C=-\frac{1}{10}$

Constant term$:3C-10D=0\Rightarrow 3\left(-\frac{1}{10}\right)-10D=0\Rightarrow D=-\frac{3}{100}$

$A=-\frac{1}{6},B=-\frac{5}{36},C=-\frac{1}{10},D=-\frac{3}{100}$

So,$y_p=-\frac{1}{6}x{e}^x-\frac{5}{36}{e}^x-\frac{1}{10}x-\frac{3}{100}$

Thus, the general solution is,$y=C_1{e}^{-5x}+C_2{e}^{2x}-\frac{1}{6}x{e}^x-\frac{5}{36}{e}^x-\frac{1}{10}x-\frac{3}{100}\text{.}$