Example 1: Separable Differential Equations-Exponential Model

Find the particular solution for the differential equation $\frac{\mathit{dy}}{\mathit{dx}}=–5y$ with the initial condition $y\mathrm{(0)}=7$.

Identifier: You are being asked to “verify that a function” is a “solution to the differential equation.

Step 1: Identify your two variables.

In this example the two variables are the y-variable and x-variable . Keep in mind even if you don’t have an x-variable visible, because the differential equation is $\frac{dy}{dx}$, you have both a y-variable and x-variable involved in the equation.

$\frac{dy}{dx}=–5y$

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

1) Multiply both sides of the equation by $\frac{1}{y}$ 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 $dx$.

$\frac{dy}{dx}=–5y$

1)

$\frac{1}{y}\bullet \frac{dy}{dx}=–5y\bullet \frac{1}{y}$

2)

$\mathit{dx}\bullet \frac{1}{y}\bullet \frac{dy}{dx}=–5\bullet \mathit{dx}$

$\frac{1}{y}\bullet dy=–5\bullet dx$

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

Keep in mind that you only need a single $\mathit{+}\mathit{C}$ that works for the entire equation, and that the best place for that $\mathit{+}\mathit{C}$ is usually on the integral that involves the x ’s. The right equation in this example.

$\int \frac{1}{y}\bullet dy=\int –5\bullet dx$

$\int \frac{1}{y}\bullet dy$

$\int –5\bullet dx$

$\ln \left(y\right)$

$–5x\mathit{+}\mathit{C}$

$\ln \mathrm{(y)}=–5x\mathit{+}\mathit{C}$

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

1) To cancel out the natural log on the left and gain access to the y -variable you will need to raise both sides to the power of $e^{\left( \right)}$. That will allow you to cancel out the natural log on the left side of the equals, freeing up your y -variable.

2) You will then want to use the laws of exponents to adjust the right side of the equals. Remembering that when you have like bases you add exponents. You will essentially be undoing that rule to break the equation back into two pieces.

3) You can then consider the $e^{\left(\mathit{C}\right)}$portion as just a constant, $\mathit{C}$ . Then move it to the front of the term where it makes more math sense.

$\ln \mathrm{(y)}=–5x\mathit{+}\mathit{C}$

1)

$e^{\mathrm{(}\ln \mathrm{(}y\mathrm{)} \mathrm{)}}=e^{\mathrm{(}–5x\mathit{+}\mathit{C} \mathrm{)}}$

$y=e^{(–5x\mathit{+}\mathit{C} )}$

2)

$y=e^{(–5x )}{e}^{\left(\mathit{C}\right)}$

3)

$y=\mathit{C}\mathit{\bullet }{e}^{(–5x )}$

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 $y\mathrm{(}0\mathrm{)}=7$. Which means that when $x=0$, $y=7$.

1) Plug in 0 for x and 7 for y in the general solution from Step 4, and evaluate any parts of the equation you can. Here you can simplify $e^{(–5(0\left) \right)}=1$

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

$y=\mathit{C}\mathit{\bullet }{e}^{(–5x )}$

$7=\mathit{C}\mathit{\bullet }{e}^{(–5(0\left) \right)}$

$7=\mathit{C}\mathit{\bullet }1$

$7=\mathit{C}$

$y=\mathit{7}\mathit{\bullet }{e}^{(–5x )}$

Final Result:

The particular solution to the differential equation $\frac{\mathit{dy}}{\mathit{dx}}=–5y$, given $y\mathrm{(}0\mathrm{)}=7$ would be $y=7\bullet e^{(–5x )}$ .