Solve the initial value problem y''' - 5y'' + 8y' - 4y =…
Solve the initial value problem $y''' - 5y'' + 8y' - 4y = e^{2x} + e^{3x}$, with $y(0)=1, \qquad y'(0)=1, \qquad y''(0)=0$.
Explanation of Method
A repeated root $r=2$ matches one exponential forcing term exactly, requiring a resonant $x^2e^{2x}$ ansatz alongside the plain $e^{3x}$ response.
Step 1: Homogeneous solution
The characteristic equation is $r^3-5r^2+8r-4=0$. Testing $r=1$: $1-5+8-4=0$, so $(r-1)$ is a factor: $r^3-5r^2+8r-4 = (r-1)(r^2-4r+4) = (r-1)(r-2)^2$.
Roots: $r=1$ (simple), $r=2$ (double). So
\[ y_h = C_1e^{x} + (C_2+C_3x)e^{2x}. \]
Step 2: Particular solution — splitting the forcing term
Write $L[y]=y'''-5y''+8y'-4y$ and split $g(x)=e^{2x}+e^{3x}$, treating each exponential separately.
Forcing term $e^{2x}$: $r=2$ is a double root.
Since $2$ is a root of multiplicity $2$, try $y_{p1}=Ax^2e^{2x}$. A direct computation gives $L[x^2e^{2x}]=2e^{2x}$ (the $x^2$ and $x$ terms cancel identically, since $r=2$ kills the polynomial part up to and including degree $1$), so $2A=1$ and $A=\tfrac12$:
\[ y_{p1} = \frac{x^2}{2}e^{2x}. \]
Forcing term $e^{3x}$: $r=3$ is not a root.
Try $y_{p2}=Be^{3x}$. Then $L[y_{p2}]=P(3)Be^{3x}$, where $P(r)=r^3-5r^2+8r-4$, and $P(3)=27-45+24-4=2$. So $2B=1$, $B=\tfrac12$: $y_{p2} = \frac{1}{2}e^{3x}$.
Hence $y_p = y_{p1}+y_{p2} = \dfrac{x^2}{2}e^{2x} + \dfrac12e^{3x}$, and the general solution of the ODE is
\[ y = C_1e^x + (C_2+C_3x)e^{2x} + \frac{x^2}{2}e^{2x} + \frac12e^{3x}. \]
Step 3: Values of $y_p$ and its derivatives at $x=0$
For $f(x)=\tfrac{x^2}{2}e^{2x}$: $f(0)=0$, $f'(x)=e^{2x}(x+x^2)$ so $f'(0)=0$, and $f''(x)=e^{2x}(2x^2+4x+1)$ so $f''(0)=1$. For $\tfrac12e^{3x}$, the values at $x=0$ are $\tfrac12$, $\tfrac32$, $\tfrac92$. Adding,
\[ y_p(0) = \frac12, \qquad y_p'(0) = \frac32, \qquad y_p''(0) = 1+\frac92 = \frac{11}{2}. \]
Step 4: Applying the initial conditions
Differentiating $y_h=C_1e^x+(C_2+C_3x)e^{2x}$ twice and evaluating at $x=0$ gives $y_h(0)=C_1+C_2$, $y_h'(0)=C_1+2C_2+C_3$, $y_h''(0)=C_1+4C_2+4C_3$. Combined with the $y_p$-values above, the conditions $y(0)=1$, $y'(0)=1$, $y''(0)=0$ become
\[ C_1+C_2 = \frac12, \qquad C_1+2C_2+C_3 = -\frac12, \qquad C_1+4C_2+4C_3 = -\frac{11}{2}. \]
Subtracting the first equation from the second: $C_2+C_3=-1$. Subtracting the second from the third: $2C_2+3C_3=-5$. Solving this $2\times2$ system: from $C_2=-1-C_3$, substitute to get $2(-1-C_3)+3C_3=-5$, i.e. $C_3=-3$, and then $C_2=2$, $C_1=\tfrac12-C_2=-\tfrac32$.
Step 5: Solution of the IVP
The solution is recorded in the result below.
Step 6: Verification
At $x=0$: $y(0)=-\tfrac32+2+\tfrac12=1.\ \checkmark$ Differentiating the boxed expression, the $e^{2x}$-block $\left(2-3x+\tfrac{x^2}2\right)e^{2x}$ has derivative $\left(-3+x+2(2-3x+\tfrac{x^2}2)\right)e^{2x}=\left(1-5x+x^2\right)e^{2x}$, so
\[ y' = -\frac32e^x + (1-5x+x^2)e^{2x} + \frac32e^{3x}, \qquad y'(0) = -\frac32+1+\frac32 = 1.\ \checkmark \]
Differentiating once more, the $e^{2x}$-block $(1-5x+x^2)e^{2x}$ has derivative $(-5+2x+2(1-5x+x^2))e^{2x}=(-3-8x+2x^2)e^{2x}$, so
\[ y'' = -\frac32e^x + (-3-8x+2x^2)e^{2x} + \frac92e^{3x}, \qquad y''(0) = -\frac32-3+\frac92 = 0.\ \checkmark \]
All three initial conditions hold, and $y$ satisfies the ODE because $y_h$ is the fundamental system at roots $1,2,2$ and $y_p$ was built term-by-term to reproduce $e^{2x}+e^{3x}$ under $L$.
Result
Final answer:
\[\boxed{\,y = -\frac{3}{2}e^{x} + \left(2-3x+\frac{x^2}{2}\right)e^{2x} + \frac{1}{2}e^{3x}\,}\]