Question

Solve the differential equation$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{cos}\left(3t\right)$and find all the real solutions of the differential equation.

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{13}\left(3\sin 3\mathrm{t}-2\cos 3\mathrm{t}\right)+{\mathrm{Ce}}^{2\mathrm{t}}$.

Step-by-step solution

Definition of first order linear differential equation

Consider the differential equation$\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{‐}\mathbf{af}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ where g(t) is a smooth function and 'a' is a constant. Then the general solution will be $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$.

Determination of the solution

Consider the differential equation as follows.

$\frac{\mathrm{dx}}{\mathrm{dt}}‐2\mathrm{x}=\cos \left(3\mathrm{t}\right)$

Now, the differential equation is in the form as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)‐\mathrm{af}\left(\mathrm{t}\right)=\mathrm{g}\left(\mathrm{t}\right)$, where g(t) is a smooth function, then the general solution will be as follows.

role="math" localid="1661140455682" $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Compute the calculation of the solution.

Substitute the value $\cos \left(3\mathrm{t}\right)$for g(t) and 2for a in$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$as follows.

localid="1661141409606" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{-2\mathrm{t}}\times \cos \left(3\mathrm{t}\right)\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt} \end{gathered}$$

Using substitution method, solve the integrall $=\int e^{-2t}\cos \left(3\mathrm{t}\right)\mathrm{dt}$by integration by parts as follows.

localid="1661141278535" $$\begin{gathered} \mathrm{u}=\cos 3\mathrm{t} \\ \mathrm{du}=‐3\sin 3\mathrm{tdt} \\ \int \mathrm{dv}=\int {\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt} \\ \mathrm{v}=\frac{{\mathrm{e}}^{-2\mathrm{t}}}{‐2} \end{gathered}$$

Substitute the value $‐3\sin 3\mathrm{tdt}$for du and$\frac{{\mathrm{e}}^{‐2\mathrm{t}}}{‐2}$for v in localid="1661141489347" $\int \mathrm{udv}=\mathrm{uv}‐\int \mathrm{vdu}$as follows.

localid="1661141523929" $$\begin{gathered} \int \mathrm{udv}=\mathrm{uv}‐\int \mathrm{vdu} \\ =‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\int ‐\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}\left(‐3\sin \left(3\mathrm{t}\right)\mathrm{dt}\right) \\ =‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int ‐{\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt} \\ \mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int ‐{\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt} \end{gathered}$$

Similarly simplify the integral as follows.

$\int {\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}$

Again using integration by parts as follows.

$$\begin{gathered} \mathrm{u}=\sin 3\mathrm{t} \\ \mathrm{du}=‐3\cos 3\mathrm{tdt} \\ \int \mathrm{dv}=\int {\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt} \\ \mathrm{v}=\frac{{\mathrm{e}}^{-2\mathrm{t}}}{‐2} \end{gathered}$$

Substitute the value$3\cos 3\mathrm{tdt}$for du and $\frac{{\mathrm{e}}^{-2\mathrm{t}}}{‐2}$for v in$\int \mathrm{udv}=\mathrm{uv}‐\int \mathrm{vdu}$as follows.

$$\begin{gathered} \int \mathrm{udv}=\mathrm{uv}‐\int \mathrm{vdu} \\ =‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\int ‐\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}\left(‐3\cos \left(3\mathrm{t}\right)\mathrm{dt}\right) \\ =‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int ‐{\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt} \\ =‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int ‐{\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt} \end{gathered}$$

Simplify further as follows.

localid="1661141863636" $$\begin{gathered} \int {\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}=‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{2}\int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt} \\ \int {\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}=‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{2} \end{gathered}$$

Substitute the valuelocalid="1661142370664" $‐\sin 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{2}\mathrm{I}$for$\int e^{-2t}\sin \left(3\mathrm{t}\right)\mathrm{dt}$in localid="1661142353137" $\mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int {\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}$as follows.

$$\begin{gathered} \mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\int {\mathrm{e}}^{-2\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt} \\ \mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}‐\frac{3}{2}\left(‐\sin \left(3\mathrm{t}\right)\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{2}\mathrm{I}\right) \\ \mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{2}\sin \left(3\mathrm{t}\right)\frac{9}{4}\mathrm{I} \\ \mathrm{I}‐\frac{9}{4}\mathrm{I}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{9}{4}\sin \left(3\mathrm{t}\right) \end{gathered}$$

Simplify further as follows.

localid="1661142642606" $$\begin{gathered} \mathrm{I}‐\frac{9}{4}\mathrm{I}=‐\cos 3t\frac{e^{-2t}}{2}+\frac{3}{4}\sin 3t \\ \frac{4\mathrm{I}‐9\mathrm{I}}{4}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{4}\sin 3\mathrm{t} \\ \frac{13\mathrm{I}}{4}=‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{4}\sin 3\mathrm{t} \\ \mathrm{I}=\frac{4}{13}\left(‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{4}\sin 3\mathrm{t}\right) \end{gathered}$$

Conclusion of the solution

Simplify further as follows.

$$\begin{gathered} \mathrm{I}=\frac{4}{13}\left(‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{3}{4}\sin 3\mathrm{t}\right) \\ \int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt}=\frac{4}{13}‐\cos 3\mathrm{t}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{2}+\frac{4}{13}\frac{3}{4}\sin 3\mathrm{t} \\ \int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt}=\frac{2}{13}‐\cos 3{\mathrm{te}}^{-2\mathrm{t}}+\frac{3}{13}\sin 3\mathrm{t} \\ \int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt}=\frac{1}{13}{\mathrm{e}}^{-2\mathrm{t}}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+\mathrm{C} \end{gathered}$$

Substitute the value $\int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt}=\frac{1}{13}{\mathrm{e}}^{-2\mathrm{t}}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+\mathrm{C}$in as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{-2\mathrm{t}}\cos \left(3\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\left(\frac{1}{13}{\mathrm{e}}^{-2\mathrm{t}}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+\mathrm{C}\right) \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{1}{13}{\mathrm{e}}^{2\mathrm{t}}{\mathrm{e}}^{-2\mathrm{t}}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{1}{13}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+{\mathrm{Ce}}^{2\mathrm{t}} \end{gathered}$$

Hence, the solution for the linear differential equation$\frac{\mathrm{dx}}{\mathrm{dt}}‐2\mathrm{x}=\cos \left(3\mathrm{t}\right)$ is $\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{13}\left(3\sin 3\mathrm{t}‐2\cos 3\mathrm{t}\right)+{\mathrm{Ce}}^{2\mathrm{t}}$.