Answer to Question #199970 in Differential Equations for sania khan

Question

solve the undetermined coefficient y''+3y'+2y=e^-x +e^-2x -x

Answer 1

Given the equation:

$y''+3y'+2y=e^{-x} +e^{-2x} -x$

We can rewrite the equation using the notation as:

$\frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-x}+e^{-2 x}-x \\$

The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving:

$\frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=0 :$

$\text{Assume a solution will be proportional to } e^{\lambda x} \text{ for some constant } \lambda.\\ \text{Substitute } y(x)=e^{\lambda x} \text{ into the differential equation:}\\$

$\frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)+3 \frac{d}{d x}\left(e^{\lambda x}\right)+2 e^{\lambda x}=0\\ \text{Substitute } \frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)=\lambda^{2} e^{\lambda x} \text{ and } \frac{d}{d x}\left(e^{\lambda x}\right)=\lambda e^{\lambda x}:\\ \lambda^{2} e^{x x}+3 \lambda e^{\lambda x}+2 e^{\lambda x}=0\\ \text{Factor out } e^{\lambda x} : \left(\lambda^{2}+3 \lambda+2\right) e^{\lambda x}=0$

$\text{Since } e^{\lambda x} \neq 0 \text{for any finite }\lambda, \text{the zeros must come from the polynomial:}\\ \lambda^{2}+3 \lambda+2=0\\ Factor:\\ (\lambda+1)(\lambda+2)=0\\ Solve for \lambda :\\ \lambda=-2 or \lambda=-1\\ The root \lambda=-2 gives y_{1}(x)=c_{1} e^{-2 x} as a solution, where c_{1} is an arbitrary constant.\\ The root \lambda=-1 gives y_{2}(x)=c_{2} e^{-x} as a solution, where c_{2} is an arbitrary constant.\\ The general solution is the sum of the above solutions:\\ y(x)=y_{1}(x)+y_{2}(x)=c_{1} e^{-2 x}+c_{2} e^{-x}\\ Determine the particular solution to\\ \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-x}+e^{-2 x}-x \\\text{by the method of undetermined coefficients: }\\ The particular solution will be the sum of the particular solutions to\\ \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=-x, \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-2 x} and\\ \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-x}$

$\text{ The particular solution to } \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=-x \text{ is of the form:}\\ y_{p_{1}}(x)=a_{1}+a_{2} x\\ \text{ The particular solution to }\frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-2 x} is of the form:\\ y_{p_{2}}(x)=x\left(a_{3} e^{-2 x}\right), where a_{3} e^{-2 x} \text{ was multiplied by} x \text{ to account for }e^{-2 x} in the complementary solution.\\ The particular solution to \frac{d^{2} y(x)}{d x^{2}}+3 \frac{d y(x)}{d x}+2 y(x)=e^{-x} is of the form:\\ y_{p_{3}}(x)=x\left(a_{4} e^{-x}\right), where a_{4} e^{-x} was multiplied by x to account for e^{-x} in the complementary solution.\\ Sum y_{p_{1}}(x), y_{p_{2}}(x), and y_{p_{3}}(x) to obtain y_{p}(x) :\\ y_{p}(x)=y_{p_{1}}(x)+y_{p_{2}}(x)+y_{p_{3}}(x)=a_{1}+a_{2} x+a_{3} e^{-2 x} x+a_{4} e^{-x} x\\ Solve for the unknown constants a_{1}, a_{2}, a_{3}, and a_{4} :\\ Compute \frac{d y_{p}(x)}{d x}:\\ \frac{d y_{p}(x)}{d x}=\frac{d}{d x}\left(a_{1}+a_{2} x+a_{3} e^{-2 x} x+a_{4} e^{-x} x\right)\\ =a_{2}+a_{3} e^{-2 x}-2 a_{3} e^{-2 x} x+a_{4} e^{-x}-a_{4} e^{-x} x$

$Compute \frac{d^{2} y_{p}(x)}{d x^{2}}:\\ \frac{d^{2} y_{p}(x)}{d x^{2}}=\frac{d^{2}}{d x^{2}}\left(a_{1}+a_{2} x+a_{3} e^{-2 x} x+a_{4} e^{-x} x\right)\\ =a_{3}\left(-4 e^{-2 x}+4 e^{-2 x} x\right)+a_{4}\left(-2 e^{-x}+e^{-x} x\right)\\ Substitute the particular solution y_{p}(x) into the differential equation:\\ \frac{d^{2} y_{p}(x)}{d x^{2}}+3 \frac{d y_{p}(x)}{d x}+2 y_{p}(x)=e^{-x}+e^{-2 x}-x\\ a_{3}\left(-4 e^{-2 x}+4 e^{-2 x} x\right)+a_{4}\left(-2 e^{-x}+e^{-x} x\right)+ 3\left(a_{2}+a_{3} e^{-2 x}-2 a_{3} e^{-2 x} x+a_{4} e^{-x}-a_{4} e^{-x} x\right)+ 2\left(a_{1}+a_{2} x+a_{3} e^{-2 x} x+a_{4} e^{-x} x\right)=e^{-x}+e^{-2 x}-x\\ Simplify:\\ 2 a_{1}+3 a_{2}-a_{3} e^{-2 x}+a_{4} e^{-x}+2 a_{2} x=e^{-2 x}+e^{-x}-x\\ Equate the coefficients of 1 on both sides of the equation:\\ 2 a_{1}+3 a_{2}=0\\ Equate the coefficients of e^{-2 x} on both sides of the equation:\\ -a_{3}=1$

$Equate the coefficients of e^{-x} on both sides of the equation:\\ a_{4}=1\\ Equate the coefficients of x on both sides of the equation:\\ 2 a_{2}=-1\\ Solve the system:\\ a_{1}=\frac{3}{4}\\ a_{2}=-\frac{1}{2}\\ a_{3}=-1\\ a_{4}=1\\ Substitute a_{1}, a_{2}, a_{3}, \text{ and }a_{4} into y_{p}(x)=a_{1}+x a_{2}+e^{-2 x} x a_{3}+e^{-x} x a_{4} :\\ y_{p}(x)=-\frac{x}{2}+e^{-x} x-e^{-2 x} x+\frac{3}{4}\\ The general solution is:\\ y(x)=y_{\mathrm{c}}(x)+y_{p}(x)=-\frac{x}{2}+e^{-x} x-e^{-2 x} x+c_{1} e^{-2 x}+c_{2} e^{-x}+\frac{3}{4}$

which is the required answer