Question: Solve $$y'' + 3y' - 4y = 6e^{2x} - 3$$, with $$y(1.5) = 4$$, $$y'(1.5) = 5$$.
1. **State the problem:** Solve the second-order linear nonhomogeneous differential equation $$y'' + 3y' - 4y = 6e^{2x} - 3$$ with initial conditions $$y(1.5) = 4$$ and $$y'(1.5) = 5$$.
2. **Find the complementary solution (homogeneous equation):** Solve $$y'' + 3y' - 4y = 0$$.
The characteristic equation is $$r^2 + 3r - 4 = 0$$.
3. **Solve the characteristic equation:**
$$r = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}$$.
So the roots are $$r_1 = 1$$ and $$r_2 = -4$$.
4. **Write the complementary solution:**
$$y_c = C_1 e^{x} + C_2 e^{-4x}$$.
5. **Find a particular solution:** Since the right side is $$6e^{2x} - 3$$, try a particular solution of the form:
$$y_p = Ae^{2x} + B$$.
6. **Compute derivatives:**
$$y_p' = 2Ae^{2x}$$
$$y_p'' = 4Ae^{2x}$$.
7. **Substitute into the differential equation:**
$$y_p'' + 3y_p' - 4y_p = 6e^{2x} - 3$$
Substitute:
$$4Ae^{2x} + 3(2Ae^{2x}) - 4(Ae^{2x} + B) = 6e^{2x} - 3$$
Simplify:
$$4Ae^{2x} + 6Ae^{2x} - 4Ae^{2x} - 4B = 6e^{2x} - 3$$
Combine like terms:
$$(4A + 6A - 4A)e^{2x} - 4B = 6e^{2x} - 3$$
$$6Ae^{2x} - 4B = 6e^{2x} - 3$$
8. **Equate coefficients:**
For $$e^{2x}$$ terms:
$$6A = 6 \implies A = 1$$
For constants:
$$-4B = -3 \implies B = \frac{3}{4}$$
9. **Write the general solution:**
$$y = y_c + y_p = C_1 e^{x} + C_2 e^{-4x} + e^{2x} + \frac{3}{4}$$
10. **Apply initial conditions:**
Calculate $$y'(x)$$:
$$y' = C_1 e^{x} - 4C_2 e^{-4x} + 2e^{2x}$$
Use $$y(1.5) = 4$$:
$$C_1 e^{1.5} + C_2 e^{-6} + e^{3} + \frac{3}{4} = 4$$
Use $$y'(1.5) = 5$$:
$$C_1 e^{1.5} - 4C_2 e^{-6} + 2e^{3} = 5$$
11. **Solve the system:**
Let $$a = e^{1.5}$$, $$b = e^{-6}$$, $$c = e^{3}$$.
Equations:
$$C_1 a + C_2 b = 4 - c - \frac{3}{4}$$
$$C_1 a - 4 C_2 b = 5 - 2 c$$
12. **Subtract first from second:**
$$(C_1 a - 4 C_2 b) - (C_1 a + C_2 b) = (5 - 2 c) - (4 - c - \frac{3}{4})$$
Simplify left:
$$C_1 a - 4 C_2 b - C_1 a - C_2 b = -5 C_2 b$$
Simplify right:
$$5 - 2 c - 4 + c + \frac{3}{4} = 1 - c + \frac{3}{4} = \frac{7}{4} - c$$
So:
$$-5 C_2 b = \frac{7}{4} - c \implies C_2 = \frac{c - \frac{7}{4}}{5 b}$$
13. **Find $$C_1$$:**
From first equation:
$$C_1 a = 4 - c - \frac{3}{4} - C_2 b$$
Substitute $$C_2$$:
$$C_1 a = 4 - c - \frac{3}{4} - b \cdot \frac{c - \frac{7}{4}}{5 b} = 4 - c - \frac{3}{4} - \frac{c - \frac{7}{4}}{5}$$
Simplify:
$$C_1 a = \left(4 - \frac{3}{4}\right) - c - \frac{c}{5} + \frac{7}{20} = \frac{13}{4} - c - \frac{c}{5} + \frac{7}{20}$$
Combine constants:
$$\frac{13}{4} + \frac{7}{20} = \frac{65}{20} + \frac{7}{20} = \frac{72}{20} = \frac{18}{5}$$
Combine $$c$$ terms:
$$-c - \frac{c}{5} = -\frac{6c}{5}$$
So:
$$C_1 a = \frac{18}{5} - \frac{6c}{5} = \frac{18 - 6c}{5}$$
Therefore:
$$C_1 = \frac{18 - 6c}{5 a} = \frac{6(3 - c)}{5 a}$$
14. **Final solution:**
$$y = \frac{6(3 - e^{3})}{5 e^{1.5}} e^{x} + \frac{e^{3} - \frac{7}{4}}{5 e^{-6}} e^{-4x} + e^{2x} + \frac{3}{4}$$