Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5

Theorem

The second order ODE:

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

with initial conditions:

\(\ds \map y 1\)

\(=\)

\(\ds 3\)

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

\(=\)

\(\ds 5\)

has the particular solution:

$y = x + 2 x^2$

Proof

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

$y = C_1 x + C_2 x^2$

Differentiating with respect to $x$:

$y' = C_1 + 2 C_2 x$

Thus for the initial conditions:

\(\ds \map y 1\)

\(=\)

\(\ds C_1 \times 1 + C_2 \times 1^2\)

\(\ds \)

\(=\)

\(\ds 3\)

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

\(\ds \leadsto \ \ \)

\(\ds C_1 + C_2\)

\(=\)

\(\ds 3\)

and:

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

\(=\)

\(\ds C_1 + 2 C_2 \times 1\)

\(\ds \)

\(=\)

\(\ds 5\)

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

\(\ds \leadsto \ \ \)

\(\ds C_1 + 2 C_2\)

\(=\)

\(\ds 5\)

\(\ds \leadsto \ \ \)

\(\ds C_2\)

\(=\)

\(\ds 2\)

$(3) - (2)$

\(\ds \leadsto \ \ \)

\(\ds C_1 + 2\)

\(=\)

\(\ds 3\)

substituting for $C_2$ in $(2)$

\(\ds \leadsto \ \ \)

\(\ds C_1\)

\(=\)

\(\ds 1\)

Hence the result.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.15$: Problem $2$