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

Jasonbradley

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Jun 16, 2024

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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_c)

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_c = c₁e^0x + c₂e^-x + c₃xe^-x

Simplifying, we get:

y_c = c₁ + c₂e^-x + c₃xe^-x

2. Finding the Particular Solution (y_p)

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_p1 = (Ax^2 + Bx + C)e^2x

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

y_p1 = (Ax^4 + Bx^3 + Cx^2)e^2x

Now we differentiate y_p1 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_p2 = 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_p2 = Dx + Ecos(2x) + Fsin(2x)

We then differentiate y_p2 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_c + y_p1 + y_p2

y = c₁ + c₂e^-x + c₃xe^-x + (Ax^4 + Bx^3 + Cx^2)e^2x + 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 c₁, c₂, c₃, 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.