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Y' + 2y = e^x

2 y{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} = e^{x}

2 y{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} = e^{x}
The differential equation.
\text{order } 1
Order 1: the highest derivative present.
\text{Exact}
M dx + N dy = 0 with ∂M/∂y = ∂N/∂x: find the potential F with F_x = M, F_y = N; the solution is F = C.
r + 2 = 0
Characteristic equation of the homogeneous part (substitute y = e^{rx}).
r = -2
Its roots.
y{\left(x \right)} = C_{1} e^{- 2 x} + \frac{e^{x}}{3}
General solution (C₁, C₂ … are arbitrary constants).
\checkmark
Verified: substituting the solution back into the equation gives 0.

Avslöja svaret

y{\left(x \right)} = C_{1} e^{- 2 x} + \frac{e^{x}}{3}