Linear First Order ODE/y' + 2 x y = exp -x^2
Jump to navigation
Jump to search
Theorem
- $\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$
has the general solution:
- $y = \paren {x + C} \map \exp {-x^2}$
Proof
$(1)$ is in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
- $\map P x = 2 x$
- $\map Q x = \map \exp {-x^2}$
Thus:
\(\ds \int \map P x \rd x\) | \(=\) | \(\ds \int 2 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int P \rd x}\) | \(=\) | \(\ds \map \exp {x^2}\) |
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
- $\map {\dfrac {\d} {\d x} } {\map \exp {x^2} y} = \map \exp {-x^2} \map \exp {x^2} = 1$
Hence:
\(\ds \map \exp {x^2} y\) | \(=\) | \(\ds \int 1 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x + C\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \paren {x + C} \map \exp {-x^2}\) |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(6)$
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $19$