Linear First Order ODE/y' = x + y/y(0) = 1

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Theorem

The linear first order ODE:

$(1): \quad \dfrac {\mathrm d y} {\mathrm d x} = x + y$

with initial condition:

$y \left({0}\right) = 1$

has the particular solution:

$y = 2 e^x - x - 1$


Proof

From Linear First Order ODE: $y' = x + y$, the general solution of $(1)$ is:

$y = C e^x - x - 1$

Setting $y = 1$ when $x = 0$ gives:

$1 = C + 1$

from which $C = 2$.

Hence the result.

$\blacksquare$


Sources