(d^3+2d^2+d)y=x^2e^2x+sin^2x

4 min read Jul 03, 2024

Solving the Differential Equation (d^3+2d^2+d)y=x^2e^2x+sin^2x

In this article, we will discuss the solution to the differential equation:

(d^3+2d^2+d)y=x^2e^2x+sin^2x

This is a third-order linear differential equation with variable coefficients. To solve it, we will use the method of undetermined coefficients.

The homogeneous equation is:

(d^3+2d^2+d)y=0

Let's assume that the solution has the form:

y_h=e^(rx)

Substituting this into the homogeneous equation, we get:

r^3+2r^2+r=0

Factoring the cubic equation, we get:

(r+1)^2(r-1)=0

This yields three roots:

r1=-1, r2=-1, r3=1

Therefore, the homogeneous solution is:

y_h=c1e^(-x)+c2xe^(-x)+c3e^(x)

The particular solution will have the form:

y_p=x^2e^2x+sin^2x

To find the coefficients, we will use the method of undetermined coefficients. Let's assume that:

y_p=Axe^2x+Bx^2e^2x+Csin^2x+Dxcos^2x

Substituting this into the original equation, we get:

(d^3+2d^2+d)(Axe^2x+Bx^2e^2x+Csin^2x+Dxcos^2x)=x^2e^2x+sin^2x

Equating the coefficients of each term, we get a system of linear equations:

A+2B+C=1 2A+4B+D=0 B-2C=0 A-2D=0

Solving this system, we get:

A=1/3, B=1/6, C=1/12, D=1/6

Therefore, the particular solution is:

y_p=(1/3)xe^2x+(1/6)x^2e^2x+(1/12)sin^2x+(1/6)xcos^2x

The general solution is the sum of the homogeneous solution and the particular solution:

y=y_h+y_p =c1e^(-x)+c2xe^(-x)+c3e^(x)+(1/3)xe^2x+(1/6)x^2e^2x+(1/12)sin^2x+(1/6)xcos^2x

This is the general solution to the differential equation (d^3+2d^2+d)y=x^2e^2x+sin^2x.

Note that the constants c1, c2, and c3 can be determined by applying the initial conditions or boundary conditions of the problem.