Solve du/dt=-k[u-T_{0}-Asin(omegat)]

$$\frac { d u } { d t } = - k [ u - T _ { 0 } - A \sin ( \omega t ) ]$$

Solve for A

$$\left\{\begin{matrix}A=\frac{u-T_{0}}{\sin(t\omega )}\text{, }&\omega \neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{\omega }\\A\in \mathrm{R}\text{, }&\left(u=T_{0}\text{ and }\omega =0\right)\text{ or }\left(\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{\omega }\text{ and }u=T_{0}\right)\text{ or }k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\left(-k\right)\left(u-T_{0}-A\sin(\omega t)\right)=\frac{\mathrm{d}}{\mathrm{d}t}(u)$$

Use the distributive property to multiply -k by u-T_{0}-A\sin(\omega t).

$$-ku+T_{0}k+Ak\sin(\omega t)=\frac{\mathrm{d}}{\mathrm{d}t}(u)$$

Add ku to both sides.

$$T_{0}k+Ak\sin(\omega t)=\frac{\mathrm{d}}{\mathrm{d}t}(u)+ku$$

Subtract T_{0}k from both sides.

$$Ak\sin(\omega t)=\frac{\mathrm{d}}{\mathrm{d}t}(u)+ku-T_{0}k$$

The equation is in standard form.

$$k\sin(t\omega )A=ku-T_{0}k$$

Divide both sides by k\sin(\omega t).

$$\frac{k\sin(t\omega )A}{k\sin(t\omega )}=\frac{k\left(u-T_{0}\right)}{k\sin(t\omega )}$$

Dividing by k\sin(\omega t) undoes the multiplication by k\sin(\omega t).

$$A=\frac{k\left(u-T_{0}\right)}{k\sin(t\omega )}$$

Divide k\left(u-T_{0}\right) by k\sin(\omega t).

$$A=\frac{u-T_{0}}{\sin(t\omega )}$$

Solve for T_0

$$\left\{\begin{matrix}\\T_{0}=-A\sin(t\omega )+u\text{, }&\text{unconditionally}\\T_{0}\in \mathrm{R}\text{, }&k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\left(-k\right)\left(u-T_{0}-A\sin(\omega t)\right)=\frac{\mathrm{d}}{\mathrm{d}t}(u)$$

Use the distributive property to multiply -k by u-T_{0}-A\sin(\omega t).

$$-ku+T_{0}k+kA\sin(\omega t)=\frac{\mathrm{d}}{\mathrm{d}t}(u)$$

Add ku to both sides.

$$T_{0}k+kA\sin(\omega t)=\frac{\mathrm{d}}{\mathrm{d}t}(u)+ku$$

Subtract kA\sin(\omega t) from both sides.

$$T_{0}k=\frac{\mathrm{d}}{\mathrm{d}t}(u)+ku-kA\sin(\omega t)$$

The equation is in standard form.

$$kT_{0}=-Ak\sin(t\omega )+ku$$

Divide both sides by k.

$$\frac{kT_{0}}{k}=\frac{k\left(-A\sin(t\omega )+u\right)}{k}$$

Dividing by k undoes the multiplication by k.

$$T_{0}=\frac{k\left(-A\sin(t\omega )+u\right)}{k}$$

Divide k\left(u-A\sin(\omega t)\right) by k.

$$T_{0}=-A\sin(t\omega )+u$$