Example 2 (Particular Solution):

Determine the particular solution to the differential equation $\frac{\mathit{dy}}{\mathit{dx}}={(y–1)}^3\cos \left(\mathit{\pi x}\right)$, given $y\mathrm{(1)}=2$.

Identifier: The directions of the problem include the language “ particular 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.

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

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

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

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

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

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

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

Step 5 (If asked to find a particular solution): Determine the particular +C value for your specific situation.

In this example you were given the fact that $y\mathrm{(}1\mathrm{)}=2$. Which means that when $x=1$, $y=2$.

1) Plug in 1 for x and 2 for y in the general solution from Step 4, and evaluate any parts of the equation you can. Here you can simplify $\frac{–2}{\pi }\mathit{sin}\mathrm{(}\pi \mathrm{(}1\mathrm{)}\mathrm{)}=0$

2) Subtract one, $–1$, from both sides to isolate the part of the equation raised to the $–\frac{1}{2}$ power.

3) Raise both sides to the $–2$ power to cancel out the $–\frac{1}{2}$ power.

4) Plug the C – value back into the general solution from Step 4.

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

1)

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

$2={\mathrm{(}0\mathit{+}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1$

$2={\mathrm{(}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1$

2)

$2–1={\mathrm{(}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}+1–1$  

$1={\mathrm{(}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}$

3)

${\mathrm{(}1\mathrm{)}}^{–2}={\mathrm{(}{\mathrm{(}\mathit{C}\mathrm{)}}^{–\frac{1}{2}}\mathrm{)}}^{–2}$ 

$1=\mathit{C}$

4)

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

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

Final Result:

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