Linear First Order ODE/y' + y = 2 x exp -x + x^2

Theorem

The linear first order ODE:

$(1): \quad y' + y = 2 x e^{-x} + x^2$

has the general solution:

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

Proof

$(1)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = 1$

Thus:

\(\ds \int \map P x \rd x\)

\(=\)

\(\ds \int 1 \rd x\)

\(\ds \)

\(=\)

\(\ds x\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int P \rd x}\)

\(=\)

\(\ds e^x\)

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\dfrac \d {\d x} e^x y = 2 x + x^2 e^x$

and the general solution is:

$e^x y = x^2 + e^x \paren {x^2 - 2 x + 2} + C$

or:

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

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $2 \ \text{(d)}$