# Solve by the method of separation of variables. x' = \frac{x}{t}

## Question:

Solve by the method of separation of variables.

{eq}x' = \frac{x}{t} {/eq}

## Separation of variables

A separable differntial equation is of the form

{eq}\displaystyle x' = f(x)g(t) {/eq}

then we can solve the differential equation by

{eq}\displaystyle \int \cfrac{xdx}{f(x)} =\int g(t) dt {/eq}

The differential equation given in this problem is separable if you take {eq}\displaystyle f(x) = x \qquad g(t) = 1/t {/eq}

## Answer and Explanation:

Before we divide by x we should check to see if we are dividing by zero. If x=0 then this is clearly a solution.

Hang on to that idea. Now divide both sides by x

{eq}\displaystyle \int \cfrac{dx}{x} = \int dt/t {/eq}

{eq}\displaystyle \ln |x| = \ln |t| + C {/eq}

{eq}\displaystyle |x| = e^{\ln |t| + C} = c_2 |t| \quad c_2>0 {/eq}

So the solution is

{eq}\displaystyle x = c_3 t {/eq}

where {eq}\displaystyle c_3{/eq} can be any value. We include the value for zero because we had the starting idea that the function x=0 is a solution. The constant can be negative due to the possibilities of the absolute values.

We can test the answer.

{eq}\displaystyle (c_3t)' = (c_3t)/t \to c_3 = c_3 {/eq}

which is valid for any real constant.

#### Learn more about this topic: Separable Differential Equation: Definition & Examples

from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 1
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