Answer to Question #207782 in Differential Equations for abigail

Question

how to solve y'''+2y''+y'=10

Answer 1

Solution.

y"'+2y''+y'=10

$\lambda^3+2\lambda^2+\lambda=0$

$\lambda(\lambda^2+2\lambda+1)=0$

$\lambda_1=0,\lambda_2=-1,\lambda_3=-1.$

From here solution of homogeneous equation $y'''+2y''+y'=0$ is

$y=C_1+C_2e^{-t}+C_3te^{-t},$

where $C_1,C_2,C_3$ are some constants.

Find particular solution of equation $y'''+2y''+y'=10$ in the form $y=At+B.$

Then

$y'=A, y''=y'''=0.$

From here $A=10,B=0.$

We will have solution

$y=C_1+C_2e^{-t}+C_3te^{-t}+10t.$

Answer. $y=C_1+C_2e^{-t}+C_3te^{-t}+10t.$