Resolution of a differential equation of first order by the method of separable variables using a given condition

 The method is the same as in resolving a differential equation without a given condition. The only difference here is to find the value of k for the given condition.

Example: Solve the following differential equation and determine the interval of validity for the given solution: dy/dx = 6y²x     y(1) = 1/25

Solution

Let's separate the variables by multiplying both sides by 1/y²:

(1/y²) (dy/dx) = (1/y²) (6y²x)

                      =  6x

Multiply both sides by dx:

1/y² dy = 6xdx

Integrate both sides:

∫ 1/y² dy = ∫6xdx

y^-1/-1 = 6x²/2 + K

-1/y = 3x² + K

-1 = y(3x² + K)

y = -1 /3x² + K

Substitute y by 1/25 and x by 1:

1/25 = -1/[3(1) +K]

1/25 = -1/3 + K

3 + K = -25

k = -25-3 k = -28

Substituting k in y we have:

y = -1/3x²-28

y exists if 3x²-28 is different of zero.

Let's study the sign of  3x²-28. In order to do that we need to equal this expression to 0 and solve the equation.

3x²-28 = 0

By solving this equation we find x = -2⎷7/3 and x = 2⎷7/3

x -∞                        -2⎷7/3            2⎷7/3            +∞

y                 +            0         -           0         +

The intervals in which 3x²-28 is different of zero are:

 -∞ < x < -2⎷7/3   -2⎷7/3 < x < 2⎷7/3  and  2⎷7/3  < x < +∞

These determine the intervals of validity of the solution of the differential equation.

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