Separable Differential Equations

Definition and Solution of a Separable Differential Equation

We have seen how to check if a given solution is a solution to a differential equation.  Now we will begin our quest to find the solution.  Although most differential equations are unsolvable, there are some that do have a solution that can be found.  We begin with the simplest type.

A differential equation is called separable if it can be written as

 f(y)dy = g(x)dx

Steps To Solve a Separable Differential Equation

To solve a separable differential equation

1. Get all the y's on the left hand side of the equation and all of the x's on the right hand side.

2. Integrate both sides.

3. Plug in the given values to find the constant of integration (C)

4. Solve for y

Example:

Solve

dy/dx = y(3 - x);   y(0 )= 5

1. dy/y = (3 - x) dx

2.

ln y = 3x - x2 / 2 + C

3. ln 5 = 0 + 0 + C

C = ln 5

Exercises:

Solve the following differential equations

1. dy/dx = x/y;   y(0) = 1

2. dy/dx = x(x+1);  y(1) = 1

3. 2xy + dy/dx = x;  y(0) = 2

Finding a Particular Solution

Example

Solve the differential equation

dy            x + 1
=                   y(0)  =  4
dx              y

Solution

Notice that this is separable.  We separate the variables.

y dy  =  (x + 1) dx

Now integrate both sides to get

y2                x2
=               + x  + C
2                 2

Now use the initial condition.  That is, plug in 0 for x and 4 for y.

42                02
=               + 0  + C
2                 2

So that

C  =  8

We can write

y2                x2
=               + x  + 8
2                 2

Multiply by 2 to get

y2  =  x2 + 2x + 16

Finally take a square root to obtain

Example

Find the equation of the function that has the following properties:

Slope  =  y sin x

and passes through the point (0,2).

Solution

Since the slope is the derivative, we have

dy / dx  =  y sin x

Separating gives

dy
=  sin x dx
y

Integrating both sides gives

ln y  =  sin x + C

Since it passes through (0,2), we plug in 0 for x and 2 for y.

ln(2)  =  sin(0) + C

ln(2)  =  C

We have

ln y  =  sin x + ln(2)

exponentiation gives

y  =  esin x + ln(2)  =  esin x eln(2)

Hence

y  =  2esin x

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