Question

solve (D⁴-2D³+D²) y=e^x + x²​

Answer 1

Step-by-step explanation:

homog part is y’’’’+2*y’’’+y’’=0 so let y=exp(m*x) to get

m^4+2*m^3+m^2=0 or m^2*(m+1)^2=0 so yh(x)=a+b*x+c*exp(-x)+d*x*exp(-x) is the solution. Now look for a particular solution yp(x)=sum(f[n]*x^n,n,2,5) since n=0,1 is covered in the homog part, and 5 is needed for y’’ to match the power of x on the rhs. This gives yp(x)=-12*x^2+3*x^3-x^4/2+x^5/20 so the general solution is yg=yh+yp=

a+b*x-12*x^2+3*x^3-x^4/2+x^5/20+c*exp(-x)+d*x*exp(-x)