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

Theorem

The linear first order ODE:

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

with initial condition:

$\map y 0 = 1$

has the particular solution:

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

Proof

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

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods