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

Theorem

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} - y = x^2$

has the general solution:

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

Proof

$(1)$ is a linear first order ODE in the form:

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

where:

$\map P x = -1$

$\map Q x = x^2$

Thus:

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

\(=\)

\(\ds -\int \rd x\)

\(\ds \)

\(=\)

\(\ds -x\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds e^{-x}\)

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

\(\ds \map {\dfrac {\d} {\d x} } {y e^{-x} }\)

\(=\)

\(\ds x^2 e^{-x}\)

\(\ds \leadsto \ \ \)

\(\ds y e^{-x}\)

\(=\)

\(\ds \int x^2 e^{-x} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac {e^{-x} } {-1} \paren {x^2 - \frac {2 x} {-1} + \frac 2 {\paren {-1}^2} } + C\)

Primitive of $x^2 e^{a x}$ for $a = -1$

\(\ds \)

\(=\)

\(\ds -e^{-x} \paren {x^2 + 2 x + 2} + C\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C e^x - \paren {x^2 + 2 x + 2}\)

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $1$