d) \( \frac{d y}{d t}-t y=y \)

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Step-by-step Solution

To solve the differential equation \[ \frac{dy}{dt} - ty = y, \] we can first rearrange it into a standard form. We can rewrite the equation as: \[ \frac{dy}{dt} - ty - y = 0, \] which simplifies to: \[ \frac{dy}{dt} - (t + 1)y = 0. \] This is a first-order linear ordinary differential equation. We can solve it using an integrating factor. The integrating factor \( \mu(t) \) is given by: \[ \mu(t) = e^{\int -(t + 1) \, dt} = e^{-(\frac{t^2}{2} + t)} = e^{-\frac{t^2}{2}} e^{-t}. \] Now, we multiply the entire differential equation by the integrating factor: \[ e^{-\frac{t^2}{2}} e^{-t} \frac{dy}{dt} - e^{-\frac{t^2}{2}} e^{-t} (t + 1)y = 0. \] This can be rewritten as: \[ \frac{d}{dt} \left( e^{-\frac{t^2}{2}} e^{-t} y \right) = 0. \] Integrating both sides gives: \[ e^{-\frac{t^2}{2}} e^{-t} y = C, \] where \( C \) is a constant of integration. Solving for \( y \), we have: \[ y = C e^{\frac{t^2}{2}} e^{t} = C e^{\frac{t^2}{2} + t}. \] Thus, the general solution to the differential equation is: \[ y(t) = C e^{\frac{t^2}{2} + t}, \] where \( C \) is an arbitrary constant.

Quick Answer

The general solution to the differential equation \( \frac{d y}{d t}-t y=y \) is \( y(t) = C e^{\frac{t^2}{2} + t} \), where \( C \) is an arbitrary constant.

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