Linear First Order ODE/y' - y = e^x/y(0) = 0

Theorem

Consider the linear first order ODE:

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

subject to the initial condition:

$\map y 0 = 0$

$(1)$ has the particular solution:

$y = x e^x$

Proof

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

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

where:

$\map P x = -1$

$\map Q x = e^x$

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:

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

\(=\)

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

\(\ds \leadsto \ \ \)

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

\(=\)

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

\(\ds \)

\(=\)

\(\ds x + C\)

Primitive of Constant

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

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

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

Substituting the initial condition into $(2)$:

\(\ds 0\)

\(=\)

\(\ds e^0 \paren {0 + C}\)

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds C\)

which leads to the particular solution:

$y = x e^x$

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $2$