Solving Separable Differential Equations Examples 4

Recall from the Solving Separable Differential Equations page that if we have a separable differential equation $M(x) + N(y) \frac{dy}{dx} = 0$, then we can rewrite it as:

(1)

\begin{align} \quad \int N(y) \: dy = \int - M(x) \: dx \end{align}

Provided that these integrals can be evaluated and that they're not too difficult to do so, then we can obtain solutions for the separable differential equation.

We will now look at some examples of solving separable differential equations. In these examples, we will concern ourselves with determining the Interval of Validity, which is the largest interval for which our solution is valid that contains the initial condition given.

Example 1

Solve the differential equation $\frac{dy}{dt} = \frac{y^2 + 1}{t + 1}$.

This differential equation can be rewritten and solved for as:

(2)

\begin{align} \quad \frac{1}{y^2 + 1} \: dy = \frac{1}{t + 1} \: dt \\ \quad \int \frac{1}{y^2 + 1} \: dy = \int \frac{1}{t + 1} \: dt \\ \quad \tan^{-1} (y) = \ln \mid t + 1 \mid + C \\ \quad y = \tan (\ln \mid t + 1 \mid + C) \end{align}

Example 2

Solve the initial value problem $\frac{dy}{dt} = \frac{4 \sin 2t}{y}$ with the initial condition $y(0) = 1$.

The given differential equation is separable, and can be rewritten and solved for as:

(3)

\begin{align} \quad y \: dy = 4 \sin 2t \: dt \\ \quad \int y \: dy = \int 4 \sin 2t \: dt \\ \quad \frac{y^2}{2} = - 2\cos 2t + C \\ \quad y^2 = -4 \cos 2t + 2C \\ \quad y = \pm \sqrt{-4 \cos 2t + 2C} \end{align}

From the initial value problem, we have that $y(0) = 1$ implies that we take the positive square root from above. Plugging in the initial condition and we have that:

(4)

\begin{align} \quad 1 = \sqrt{-4 \cos (0) + 2C} \\ \quad 1 = \sqrt{-4 + 2C} \\ \quad 1 = -4 + 2C \\ \quad 5 = 2C \\ \quad C = \frac{5}{2} \end{align}

Therefore the solution to our initial value problem is:

(5)

\begin{align} \quad y = \sqrt{-4 \cos 2t + 5} \end{align}

Example 3

Solve the differential equation $\frac{dy}{dx} = \frac{y^2(x - 3)}{x^3}$.

This differential equation is separable and can be rewritten and solved as follows:

(6)

\begin{align} \quad \frac{1}{y^2} \: dy = \frac{x - 3}{x^3} \: dx \\ \quad \frac{1}{y^2} \: dy = \left ( \frac{1}{x^2} - \frac{3}{x^3} \right ) \: dx \\ \quad \int \frac{1}{y^2} \: dy = \int \left ( \frac{1}{x^2} - \frac{3}{x^3} \right ) \: dx \\ \quad -\frac{1}{y} = - \frac{1}{x} + \frac{3}{2x^2} + C \\ \quad -\frac{1}{y} = \frac{-2x + 3 + 2Cx^2}{2x^2} \\ \quad \frac{1}{y} = \frac{2x - 3 - 2Cx^2}{2x^2} \\ \quad y = \frac{2x^2}{2x - 3 - 2Cx^2} \end{align}