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Separable Differential Equations

April 5th, 2009 |

Author: Merrin

Separable differential equations, which are also first order, are of the form

$\displaystyle \cfrac{dy}{dx} =f(x)g(y)$

The method to solve separable differential equations is to put variables of x on one side of the equation and variables of y on the other side of the equal sign.

$\displaystyle \cfrac{dy}{g(y)} =f(x)dx$

Integrate each side as an indefinite integral and remember to include the combined integration constant.

$\displaystyle \int \cfrac{dy}{g(y)} =\int f(x)dx$

A final step is to solve for y as a function of x if possible, otherwise just leave the result as an implicit relationship.

Or to avoid using integration constants requires knowing

$\displaystyle x_{1} ,x_{2},y_{1},y_{2}$

In this case,

$\displaystyle \int_{y_{1} }^{y_{2} }\cfrac{dy}{g(y)}=\int_{x_{1} }^{x_{2} }f(x)dx$

is the solution.

Example 1. Solve the following differential equation

$\displaystyle \cfrac{dy}{dx} =\sin x\sqrt{1-y^{2} }$

Solution 1. We first separate the variables.

$\displaystyle \cfrac{dy}{\sqrt{1-y^{2} } } =\sin xdx$

Then we take the indefinite integral

$\displaystyle \int \cfrac{dy}{\sqrt{1-y^2}}=\int \sin xdx$

The result is

$\displaystyle \arcsin y=-\cos x+C$

It is usual to solve for y if possible. In this case, we write

$\displaystyle y=\sin (C-\cos x)$

Example 2. Solve the following differential equation

$\displaystyle \cfrac{dy}{dx} =xe^{x-y}$

Solution 2. First separate the variables,

$\displaystyle e^y dy =xe^x dx$

then take the indefinite integral of both sides

$\displaystyle \int e^y dy=\int xe^x dx$
$\displaystyle e^y =xe^x -e^x + C$

In the last step, solve for y

$\displaystyle y = \ln (xe^x - e^x + C)$

If there is an initial value problem for this differential equation we could determine the integration constant. Let us define a typical initial value problem as y(0)=1. This gives an equation for the constant in the previous solution

$\displaystyle 1=\ln (C-1)$
$C = e + 1$

Example 3. Another type of equation which technically is separable is when

$\displaystyle \cfrac{dy}{dx}=0$

Solution 3. We can integrate both the left and right hand sides of the equation with respect to $x$ and we get.

$\displaystyle \int\cfrac{dy}{dx}=\int 0dx\qquad y=C$

Another possible generalization of the previous equation is

$\displaystyle \cfrac{d^{3} y}{dx^3}=0$
$\displaystyle \cfrac{d^{2} y}{dx^{2} }=a$
$\displaystyle \cfrac{dy}{dx}=ax+b$
$\displaystyle y = \cfrac{a}{2} x^{2}+bx+c$