Example 1 (General Solution):

Determine the general solution to the differential equation $\frac{\mathit{dy}}{\mathit{dx}}={(y–1)}^3\cos \left(\mathit{\pi x}\right)$.

Identifier: The directions of the problem include the language “ general solution ” and “ differential equation ”.

Step 1: Identify your two variables.

In this example the two variables are the y-variable and x-variable .

$\frac{dy}{dx}={(y–1)}^3\cos \left(\pi x\right)$

Step 2: Separate your equation by variables on to either side of the equals.

1) Multiply both sides of the equation by $\frac{1}{{(y–1)}^3}$ to get the y-variables on the same side as $dy$.

2) Multiply both sides of the equation by $\mathit{dx}$ to get the x-variables on the same side as $dx$.

$\frac{dy}{dx}={(y–1)}^3\cos \left(\pi x\right)$

1)                $\frac{1}{{(y–1)}^3}\bullet \frac{dy}{dx}={(y–1)}^3\cos \left(\pi x\right)\bullet \frac{1}{{(y–1)}^3}$

2)                $\mathit{dx}\bullet \frac{1}{{(y–1)}^3}\bullet \frac{dy}{dx}=\cos \left(\pi x\right)\bullet \mathit{dx}$

$\frac{1}{{(y–1)}^3}\bullet dy=\cos \left(\pi x\right)\bullet dx$

Step 3: Take the antiderivative of both sides of the equation.

Both antiderivatives require you to run a u-substitution process.

Keep in mind that you only need a single $\mathit{+}\mathit{C}$ that works for the entire equation.

$\int \frac{1}{{(y–1)}^3}\bullet dy=\int \cos \left(\pi x\right)\bullet dx$

$\int \frac{1}{{(y–1)}^3}\bullet dy$

$\int \cos \left(\pi x\right)\bullet dx$

$\int \frac{1}{{(y–1)}^3}\mathit{dy}$

$u=y–1$

$\frac{du}{dy}=1$

$du=dy$

$\int \frac{1}{{\left(u\right)}^3}du$

$\int {\left(u\right)}^{–3}du$

$\int u^{–3}du=\frac{u^{–2}}{–2}$

$\frac{u^{–2}}{–2}=\frac{{(y–1)}^{–2}}{–2}$

$\int \cos \left(\mathit{\pi x}\right)\bullet \mathit{dx}$

$u=\mathit{\pi x}$

$\frac{du}{dx}=\pi$

$du=\pi \bullet dx$

$\frac{du}{\pi }=dx$

$\int \cos \left(u\right)\bullet \frac{du}{\pi }$

$\frac{1}{\pi }\int \cos \left(u\right)du$

$\frac{1}{\pi }\int \cos \left(u\right)du=\frac{1}{\pi }\mathit{sin}\left(u\right)\mathit{+}\mathit{C}$

$\frac{1}{\pi }\mathit{sin}\left(u\right)\mathit{+}\mathit{C}\mathit{=}\frac{1}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}$

$\int \frac{1}{{(y–1)}^3}\bullet dy=\int \cos \left(\pi x\right)\bullet dx$

$\frac{{(y–1)}^{–2}}{–2}=\frac{1}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}$

Step 4: Solve the equation you found in Step 3 to get the dependent variable ( y -variable) alone on one side of the equals.

$\frac{{(y–1)}^{–2}}{–2}=\frac{1}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}$

$–2\bullet \frac{{(y–1)}^{–2}}{–2}=\mathrm{(}\frac{1}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}\bullet –2$

${(y–1)}^{–2}=\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}$

${\mathrm{(}{(y–1)}^{–2}\mathrm{)}}^{–\frac{1}{2}}={\mathrm{(}\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}$

$y–1={\mathrm{(}\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}$

$y–1+1={\mathrm{(}\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1$

$y={\mathrm{(}\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1$

Final Result:

The general solution to the differential equation $\frac{\mathit{dy}}{\mathit{dx}}={(y–1)}^3\cos \left(\mathit{\pi x}\right)$ would be $y={\mathrm{(}\frac{–2}{\pi }\mathit{sin}\left(\mathit{\pi x}\right)\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1$.