Linear First Order ODE/y' = x + y
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Theorem
- $(1): \quad \dfrac {\d y} {\d x} = x + y$
has the general solution:
- $y = C e^x - x - 1$
Initial Value: $\map y 0 = 1$
- $(1): \quad \dfrac {\d y} {\d x} = x + y$
with initial condition:
- $\map y 0 = 1$
has the particular solution:
- $y = 2 e^x - x - 1$
Proof
Rearranging $(1)$:
- $(2): \quad \dfrac {\d y} {\d x} - y = x$
$(2)$ is a linear first order ODE in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
- $\map P x = -1$
- $\map Q x = x$
Thus:
\(\ds \int \map P x \rd x\) | \(=\) | \(\ds -\int \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int P \rd x}\) | \(=\) | \(\ds e^{-x}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {e^x}\) |
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
\(\ds \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }\) | \(=\) | \(\ds \frac x {e^x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac y {e^x}\) | \(=\) | \(\ds \int \frac x {e^x} \rd x\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {e^x} \paren {x + 1} + C\) | Primitive of $x e^{a x}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds C e^x - x - 1\) |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods