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

This article will guide you through solving the third-order linear non-homogeneous differential equation:

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

**1. Finding the Complementary Solution (y<sub>c</sub>)**

First, we need to find the complementary solution, which is the solution to the homogeneous equation:

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

To do this, we find the roots of the characteristic equation:

**m^3 + 2m^2 + m = 0**

Factoring out m, we get:

**m(m^2 + 2m + 1) = 0**

Further factoring, we have:

**m(m + 1)^2 = 0**

This gives us the roots:

**m = 0**(multiplicity 1)**m = -1**(multiplicity 2)

Therefore, the complementary solution is:

**y<sub>c</sub> = c<sub>1</sub>e<sup>0x</sup> + c<sub>2</sub>e<sup>-x</sup> + c<sub>3</sub>xe<sup>-x</sup>**

Simplifying, we get:

**y<sub>c</sub> = c<sub>1</sub> + c<sub>2</sub>e<sup>-x</sup> + c<sub>3</sub>xe<sup>-x</sup>**

**2. Finding the Particular Solution (y<sub>p</sub>)**

Now we need to find a particular solution to the non-homogeneous equation. Since the right-hand side consists of two terms, we will use the method of undetermined coefficients, finding a particular solution for each term separately.

**2.1 Particular Solution for x^2e^2x**

The term x^2e^2x suggests a particular solution of the form:

**y<sub>p1</sub> = (Ax^2 + Bx + C)e<sup>2x</sup>**

However, since the root m = -1 has multiplicity 2, we need to multiply this guess by x^2:

**y<sub>p1</sub> = (Ax^4 + Bx^3 + Cx^2)e<sup>2x</sup>**

Now we differentiate y<sub>p1</sub> three times and substitute it into the original non-homogeneous equation to solve for A, B, and C.

**2.2 Particular Solution for sin^2x**

For sin^2x, we can use the identity sin^2x = (1 - cos(2x))/2. This suggests a particular solution of the form:

**y<sub>p2</sub> = D + Ecos(2x) + Fsin(2x)**

However, since the constant term is already present in the complementary solution, we need to multiply this guess by x:

**y<sub>p2</sub> = Dx + Ecos(2x) + Fsin(2x)**

We then differentiate y<sub>p2</sub> three times and substitute it into the original non-homogeneous equation to solve for D, E, and F.

**3. Combining the Solutions**

The general solution to the original differential equation is the sum of the complementary solution and the particular solution:

**y = y<sub>c</sub> + y<sub>p1</sub> + y<sub>p2</sub>**

**y = c<sub>1</sub> + c<sub>2</sub>e<sup>-x</sup> + c<sub>3</sub>xe<sup>-x</sup> + (Ax^4 + Bx^3 + Cx^2)e<sup>2x</sup> + Dx + Ecos(2x) + Fsin(2x)**

**4. Solving for the Constants**

To find the specific solution, we need initial conditions or boundary conditions to solve for the constants c<sub>1</sub>, c<sub>2</sub>, c<sub>3</sub>, A, B, C, D, E, and F.

This detailed breakdown provides a step-by-step guide to solving the given differential equation. Remember, this is a complex process involving finding roots, applying undetermined coefficients, and solving a system of equations.