The problem shown in the image is a second-order linear differential equation:

$y'' + 5y' + 6y = e^{4x}.$

Solution Outline:

This is a non-homogeneous linear differential equation. The general solution is the sum of the complementary solution $y_c$ (for the homogeneous equation) and a particular solution $y_p$ for the non-homogeneous term.

1. Solve the Homogeneous Equation:

$y'' + 5y' + 6y = 0$

Find the characteristic equation: $r^2 + 5r + 6 = 0$ Factorize: $(r + 2)(r + 3) = 0 \quad \implies \quad r = -2, \, r = -3.$

The complementary solution is: $y_c = C_1 e^{-2x} + C_2 e^{-3x},$ where $C_1$ and $C_2$ are arbitrary constants.

2. Find a Particular Solution:

The non-homogeneous term is $e^{4x}$. We use the method of undetermined coefficients. Assume the particular solution has the form: $y_p = Ae^{4x}.$

Substitute $y_p = Ae^{4x}$ into the original equation: $(A(4)^2 e^{4x} + 5A(4)e^{4x} + 6Ae^{4x}) = e^{4x}.$

Simplify: $16A e^{4x} + 20A e^{4x} + 6A e^{4x} = e^{4x}.$

Combine coefficients: $(16A + 20A + 6A)e^{4x} = e^{4x}.$

$42A e^{4x} = e^{4x}.$

Solve for $A$: $A = \frac{1}{42}.$

Thus, the particular solution is: $y_p = \frac{1}{42} e^{4x}.$

3. General Solution:

The general solution is: $y = y_c + y_p = C_1 e^{-2x} + C_2 e^{-3x} + \frac{1}{42} e^{4x}.$

Would you like further explanation on any step? Let me know! 😊

Related Questions:

1. How do we determine the form of the particular solution for non-homogeneous equations?
2. What is the significance of the characteristic equation in solving differential equations?
3. Can we apply variation of parameters to solve this problem? How would it differ?
4. How do the roots of the characteristic equation affect the complementary solution?
5. Why is the undetermined coefficient method chosen here over other methods?

Tip:

Always check whether the assumed form for $y_p$ overlaps with $y_c$. If it does, multiply $y_p$ by $x$ to ensure linear independence.