To solve the differential equation

\frac{\mathrm{d}T}{\mathrm{d}t} = A\sin\left(\omega t\right) - kT

we will use the method of integrating factors. This method is useful for solving first-order linear differential equations of the form

\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)

In our case, T is the dependent variable and t is the independent variable. We can rewrite the given differential equation in the standard form:

\frac{\mathrm{d}T}{\mathrm{d}t} + kT = A\sin\left(\omega t\right)

Here, P(t) = k and Q(t) = A\sin\left(\omega t\right). The integrating factor, \mu(t), is given by :

\mu(t) = e^{\int P(t) \, \mathrm{d}t} = e^{\int k \, \mathrm{d}t} = e^{kt}

Multiplying the entire differential equation by the integrating factor:

e^{kt} \frac{\mathrm{d}T}{\mathrm{d}t} + ke^{kt}T = Ae^{kt}\sin\left(\omega t\right)

The left-hand side of this equation is the derivative of the product e^{kt}T:

\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{kt}T\right) = Ae^{kt}\sin\left(\omega t\right)

Integrating both sides with respect to t:

\int \frac{\mathrm{d}}{\mathrm{d}t}\left(e^{kt}T\right) \, \mathrm{d}t = \int Ae^{kt}\sin\left(\omega t\right) \, \mathrm{d}t

e^{kt}T = \int Ae^{kt}\sin\left(\omega t\right) \, \mathrm{d}t + C

where C is the constant of integration.

Evaluating \int Ae^{kt}\sin\left(\omega t\right)\,\mathrm{d}t

To solve the integral on the right-hand side, we use integration by parts . Let:

u=\sin\left(\omega t\right), so \mathrm{d}u = \omega \cos\left(\omega t\right) \, \mathrm{d}t
\mathrm{d}v=Ae^{kt}\,\mathrm{d}t, so v = \frac{A}{k}e^{kt}

Applying integration by parts:

\int Ae^{kt}\sin\left(\omega t\right) \, \mathrm{d}t = \frac{A}{k}e^{kt}\sin\left(\omega t\right) - \int \frac{A\omega}{k}e^{kt}\cos\left(\omega t\right) \, \mathrm{d}t

For the second integral, apply integration by parts again. Let:

u = \cos\left(\omega t\right), so \mathrm{d}u = -\omega \sin\left(\omega t\right) \, \mathrm{d}t
\mathrm{d}v = \frac{A\omega}{k}e^{kt} \, \mathrm{d}t, so v = \frac{A\omega}{k^2}e^{kt}

Applying integration by parts:

\int \frac{A\omega}{k}e^{kt}\cos\left(\omega t\right) \, \mathrm{d}t = \frac{A\omega}{k^2}e^{kt}\cos\left(\omega t\right) + \int \frac{A\omega^2}{k^2}e^{kt}\sin\left(\omega t\right) \, \mathrm{d}t

Substituting back, we have:

\int Ae^{kt}\sin\left(\omega t\right) \, \mathrm{d}t = \frac{A}{k}e^{kt}\sin\left(\omega t\right) - \left( \frac{A\omega}{k^2}e^{kt}\cos\left(\omega t\right) + \int \frac{A\omega^2}{k^2}e^{kt}\sin\left(\omega t\right) \, \mathrm{d}t \right)

ie

I=\frac{A}{k}e^{kt}\sin\left(\omega t\right)-\frac{A\omega}{k^2}e^{kt}\cos\left(\omega t\right)-\frac{\omega^2}{k^2}I

I\left(\frac{\omega^2+k^2}{k^2}\right)

Solving the resulting equation, we find:

I=\frac{A}{k^2+\omega^2}\left(k\sin\left(\omega t\right)-\omega\cos\left(\omega t\right)\right)e^{kt}

Substituting back, we have:

e^{kt}T = \frac{A}{k^2 + \omega^2} \left( k\sin\left(\omega t\right) - \omega\cos\left(\omega t\right) \right) e^{kt} + C

Dividing through by e^{kt} gives the solution for T(t):

T(t)=\frac{A}{k^2+\omega^2}\left(k\sin\left(\omega t\right)-\omega\cos\left(\omega t\right)\right)+Ce^{-kt}

Differential Equations