Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1

Theorem

The second order ODE:

$(1): \quad y - 3 y' + 2 y = 0$

with initial conditions:

$\map y 0 = -1$

$\map {y'} 0 = 1$

has the particular solution:

$y = -3 e^x + 2 e^{2 x}$

Proof

From Linear Second Order ODE: $y - 3 y' + 2 y = 0$, the general solution of $(1)$ is:

$y = C_1 e^x + C_2 e^{2 x}$

Differentiating with respect to $x$:

$y' = C_1 e^x + 2 C_2 e^{2 x}$

Thus for the initial conditions:

\(\ds \map y 0\)

\(=\)

\(\ds C_1 e^0 + C_2 e^{2 \times 0}\)

\(\ds \)

\(=\)

\(\ds -1\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds C_1 + C_2\)

\(=\)

\(\ds -1\)

and:

\(\ds \map {y'} 1\)

\(=\)

\(\ds C_1 e^0 + 2 C_2 e^{2 \times 0}\)

\(\ds \)

\(=\)

\(\ds 1\)

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds C_1 + 2 C_2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds C_2\)

\(=\)

\(\ds 2\)

$(3) - (2)$

\(\ds \leadsto \ \ \)

\(\ds C_1 + 2\)

\(=\)

\(\ds -1\)

substituting for $C_2$ in $(2)$

\(\ds \leadsto \ \ \)

\(\ds C_1\)

\(=\)

\(\ds -3\)

Hence the result.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.15$: Problem $3$