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# Solve the given Differential Equation by Undetermined Coefficients. 1/4 y'' + y' + y = x^{2} − 4x

**Solution:**

We will be solving this by finding the homogeneous and complementary solutions of the equation. Let's solve this step by step.

Given, differential equation: 1/4 y'' + y' + y = x^{2} − 4x

First solve the corresponding homogeneous differential equation: 1/4 y'' + y' + y = 0

The characteristic equation is 1/4 r^{2} + r + 1 = 0

Let's find its roots by factorization.

(1/2 r + 1) (1/2 r + 1) = 0

1/2 r + 1 = 0

r = -2

Therefore, r = -2 is a repeated root thus one of the complimentary solutions y_{c} is y_{c} = c_{1}e^{-2x} + c_{2}xe^{-2x}

Now, find the remaining complimentary solution y_{p}.

y_{p }= Ax^{2} + Bx + C

y′_{p }= 2Ax + B

y′′_{p }= 2A

Substitute y_{p}, y′_{p}, and y′′_{p} in given differential equation and solve for A and B.

1/4(2A) + (2Ax + B) + (Ax^{2} + Bx + C) = x^{2} − 4x

1/2A + 2Ax + B + Ax^{2} + Bx +C = x^{2} − 4x

Ax^{2} + x(2A + B) + (1/2A + B + C) = x^{2} − 4x

On comparing, we get A = 1, 2A + B = -4, 1/2A + B + C = 0

So, B = -6

and,

1/2A + B + C = 0

1/2 - 6 + C = 0

C = 5 -1/2

C = 11/2

Therefore, y_{p }= x^{2} - 6x + 11/2

Now, combine complementary solution and particular solution together to arrive at the general solution.

y = y_{c} + y_{p}

y = c_{1}e^{-2x} + c_{2}xe^{-2x} + x^{2} - 6x + 11/2

Thus, the general solution of differential equation 1/4 y'' + y' + y = x^{2} − 3x is y = c_{1}e^{-2x} + c_{2}xe^{-2x} + x^{2} - 6x + 11/2

# Solve the given Differential Equation by Undetermined Coefficients. 1/4 y'' + y' + y = x^{2} − 4x

**Summary:**

The general solution of differential equation 1/4 y'' + y' + y = x^{2} − 4x is y = c_{1}e^{-2x} + c_{2}xe^{-2x} + x^{2} - 6x + 11/2

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