Question

Use the method of undetermined coefficients to solve the following differential equation: a. y′′+4y=4x b.y′′−6y′+9y=324(e⁹ᵗ) with y(0)=8 and y′(0)=4. c.y′′+7y′+10y=−12t(e³ᵗ).

Answer 1

a.
`\( y(x) = C₁cos(2x) + C₂sin(2x) + x \)` where ( C₁ ) and ( C₂ ) are constants.

b.
`\( y(t) = (A + Bt)e^(3t) + 6e^(3t) \)`where ( A = 8 ) and ( B = -12 ).

c.
`\( y(t) = C₁e^(-2t) + C₂e^(-5t) - 2e^(-2t)te^(3t) \)` where ( C₁) and ( C₂) are constants.

Step-by-step explanation:

a. For the differential equation
`\(y'' + 4y = 4x\)`, we assume a particular solution in the form
`\(y_p = Ax + B\)` since the non-homogeneous term is linear. After finding the particular solution, we combine it with the complementary solution obtained from the characteristic equation to form the general solution
`\(y = C₁cos(2x) + C₂sin(2x) + x\).`

b. In the case of
`\(y'' - 6y' + 9y = 324e^(9t)\)`, the non-homogeneous term involves an exponential function. We assume a particular solution
`\(y_p = (A + Bt)e^(9t)\)`. After solving for the constants (A) and (B), combining the particular and complementary solutions results in
`\(y = (A + Bt)e^(9t) + 6e^(3t)\)`, and using the initial conditions, we find (A = 8) and (B = -12).

c. For
`\(y'' + 7y' + 10y = -12te^(3t)\)`, we assume
`\(y_p = C₁e^(-2t) + C₂e^(-5t) + De^(-2t)te^(3t)\)` as a particular solution. The general solution is then obtained by combining the particular and complementary solutions, resulting in
`\(y = C₁e^(-2t) + C₂e^(-5t) - 2e^(-2t)te^(3t)\).`