IGNOU MTE 8 2018 - Find a continuous solution of the IVPwhere g(t) = .

Question

Find a continuous solution of the IVP
$\frac{dy}{dx} + y = g(t) = 0$

where g(t) = $\left\{\begin{matrix} & \left \ 2, 0\leq t \leq 1\\ & 0, t> \gslant 1 \end{matrix}\right.$ .

13 Feb 2018

Answer :

Word Count : 469

To solve the given initial value problem (IVP): \[ \frac{dy}{dx} + y = g(t) \] with $ g(t) $ defined as: \[ g(t) = \begin{cases} 2, & 0 \leq t \leq 1 \\ 0, & t > 1 \end{cases} \] ### Step 1: Solve the equation for $ 0 \leq t \leq 1 $ (when $ g(t) = 2 $): The differential equation becomes: \[ \frac{dy}{dx} + y = 2 \] This is a first-order linear differential equation, which can be solved using the integrating factor method. The integrating factor $ \mu(t) $ is given by: \[ \mu(t) = e^{\int 1 \, dt} = e^t \] Multiplying both sides of the differential equation by $ \mu(t) $: \[ e^t \frac{dy}{dt} + e^t y = 2 e^t \] This simplifies to: \[ \frac{d}{dt} \left( e^t y \right) = 2 e^t \] Integrating both sides with respect to $ t $: \[

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