Question

Y" +2y' +5y=1 +e^x Solve the differential equation or initial-value problem using

the method of undetermined coefficients.

Answer 1

`y=e^(-x)(C_1\cos(2x)+C_2\sin(2x))+(e^x)/(8)+\frac15` or
`y=((C_1\cos(2x)+C_2\sin(2x)))/(e^x)+(e^x)/(8)+\frac15`.

Explanation:

this problem has two steps, the first is the characteristic equation and the second is the particular integral.

the characteristic equation is based on the left hand side, and we first let that equal to 0. letting
`y=e^(rx)`, we have
`y'=re^(rx), y''=r^2e^(rx)`. substituting these values we have

`r^2e^(rx)+2re^(rx)+5e^(rx)=0`, and dividing by
`e^(rx)` gives
`r^2+2r+5=0`. using the quadratic formula, we get:

`r=(-2\pm√(2^2-4(5)))/(2)=(-2\pm4i)/(2)=-1\pm2i`,

by using
`√(-1)=i`.

now, we use the general form for complex solutions:

if
`r=a\pm bi`, then
`y=e^a(C_1\cos(bx)+C_2\sin(bx))`.

here, we would have
`y=e^(-x)(C_1\cos(2x)+C_2\sin(2x))`. note that
`C_1,\ C_2` are arbitrary constants.

next, we have the particular integral. we need to add this to the value of y above. we will use undetermined coefficients here. first, considering the right hand side we have
`1+e^x`, so we let
`y=Ae^x+Bx^2+Cx+D`.

then, we have
`y'=Ae^x+2Bx+C` and
`y''=Ae^x+2B`.

substituting these values to the equation, we have

`Ae^x+2B+2Ae^x+4Bx+2C+5Ae^x+5Bx^2+5Cx+5D=1+e^x`.

simplify the equation a bit:

`8Ae^x+5Bx^2+(4B+5C)x+(2B+2C+5D)=1+e^x`.

next, compare coefficients. note that the coefficients of
`x,\ x^2` are 0.

`8A=1\implies A=\frac18\\5B=0\implies B=0\\4B+5C=0\implies 5C=0\implies C=0\\2B+2C+5D=1\implies 5D=1\implies D=\frac15`

finally, we have the particular integral as
`y=(e^x)/(8)+\frac15`.

thus, our solution is
`y=e^(-x)(C_1\cos(2x)+C_2\sin(2x))+(e^x)/(8)+\frac15`.