ODE 1st orderMedium

Linear 1st order ODE

Solve y' + 3y = 6 with y(0) = 0.

Given data

y' + 3y = 6y(0) = 0

Steps

0 / 3

Calculate the integrating factor μ = e^(∫3 dx).
Write your working — the numerical result must appear in the text (logica)
Integrate (μy)' = μ·6 = 6e^(3x).
Apply y(0) = 0 and find C.

Full worked solution

Calculate the integrating factor μ = e^(∫3 dx).
$\mu = e^{3x}$
$\mu = e^{\int 3\,dx} = e^{3x}$ . The integrating factor for the linear ODE $y' + P(x)y = Q(x)$ with $P(x) = 3$ .
Integrate (μy)' = μ·6 = 6e^(3x).
$e^{3x}y = 2e^{3x} + C$
$(\mu y)' = \mu Q = 6e^{3x} \rightarrow \mu y = \int 6e^{3x}\,dx = 2e^{3x} + C$ . Solving $y = 2 + C e^{-3x}$ .
Apply y(0) = 0 and find C.
$0 = 2 + C \Rightarrow C = -2$
$0 = 2 + C\,e^{0} \rightarrow C = -2$ . The particular solution satisfying $y(0) = 0$ is $\mathbf{y(x) = 2(1 - e^{-3x})}$ .

Result:y(x) = 2(1 − e^{−3x}).

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