how to find the solution of the ode y”-2y’+8y=e^x
Solution :
Given non homogeneous linear ode y”-2y’+8y=e^x \[y^{\prime\prime}-2y^\prime+8y=e^x\] Subject to initial conditions \[ y\left(0\right)=1,\ y^\prime\left(0\right)=-1 \] \[\frac{d^2y}{dx^2}-2\frac{dy}{dx}+8y=e^x\] Let D=d/dx \[D^2y-2Dy+8y=e^x\] \[\left(D^2-2D+8\right)y=e^x\] Step 1: To find complementary solution (yc) that is to solve \[\frac{d^2y}{dx^2}-2\frac{dy}{dx}+8y=0\] (1) So it’s auxiliary equation is obtained by substitutiting \[\frac{d}{dx}=\lambda,\ \frac{d^2}{dx^2}=\lambda^2\] In eqn (1), we obtain \[ \lambda^2+2\lambda-8=0\] \[\lambda^2+4\lambda-2\lambda-8=0\] \[\lambda\left(\lambda+4\right)-2\left(\lambda+4\right)=0\] \[\left(\lambda+4\right)\left(\lambda-2\right)=0\] It’s roots are \[\lambda_1=-4 , \lambda_2=2\] Hence the complementary solution is \[y_c(x)=c_1e^{-4x}+c_2e^{2x} \] (2) Step 2 : To find particular integral (yp) Since \[y_p\left(x\right)=\frac{e^x}{f(D)}\] where f(D)=D^2-2D+8 We know that if f(a) is not zero then \[\frac{e^{ax}}{f\left(D\right)}=\frac{e^{ax}}{f\left(a\right)}\] \[y_p\left(x\right)=\frac{e^x}{f(1)}=\frac{e^x}{1^2-2\ast1+8\ }=\frac{e^x}{7} \] (3) Step 3 : To write general solution \[y\left(x\right)=y_c\left(x\right)+y_p\left(x\right)=c_1e^{-4x}+c_2e^{2x}+\frac{e^x}{7} \] (4) Step 4 : Applying given initial conditions in eqn (4) Substituting x =0 and using first initial condition \[y\left(0\right)=1\] \[1=c_1+c_2+\frac{e}{7} \] \[1-\frac{e}{7}=c_1+c_2 \] (5) Differentiatitng eqn (4) \[y^\prime\left(x\right)=-4c_1e^{-4x}+2c_2e^{2x}+\frac{e^x}{7} \] (6) Substituting x=0 and using second initial condition \[y^\prime\left(0\right)=-1\] -1=-4c_1+2c_2+\frac{e}{7} \[ -1-\frac{e}{7}=-4c_1+2c_2\] (5) And solving eqns (5) and (6) we get \[c_1=\frac{\left(21-e\right)}{42},\ \ c_2=\frac{\left(21-5e\right)}{42}\] Substituting in the eqn (4) \[ y\left(x\right)=\frac{\left(21-e\right)}{42}e^{-4x}+\frac{\left(21-5e\right)}{42}e^{2x}+\frac{e^x}{7}\] Is the required solution of the non homogeneous differential equation.