Finding the solution of differential equations for a given condition

A differential equation has several solutions. Solving the differential equation for a given condition is equivalent to finding the solution that satisfies the condition. We apply the general process described earlier except that we find the constant c that satisfies the condition.

Example: 

Find the solution to the following differential equation: ty' + 2y = t² - t + 1 for the condition y(1) = 1/2

Solution

The general form of a differential equation of first order is : dy/dt + p(t)y = g(t).

Let's put the given equation in the general form by dividing both sides by t:
t/ty' + 2/ty = t²/t - t/t + 1/t
y' + 2/ty = t -1 + 1/t

Let's find μ(t) by substituting p(t) = 2/t in the formula:

Let's multiply the differential equation in its proper form by mu of t = t²

  (t²)y' +  (t²)2/ty = t³ -t² + t² 1/t
t²y' + 2ty = t³ -t² + t
The left side is the derivative of t²y. We have:
(t²y)' =  t³ -t² + t
Integrate both sides:
∫(t²y)'dt = ∫( t³ -t² + t)dt
t²y = t⁴/4 - t³/3 + t²/2 + c
y =  t⁴/4t² - t³/3t² + t²/2t² + c/t²
y = 1/4t² - 1/3t + 1/2 + c/t²
Let's substitute y by 1/2 and t by 1 in order to find c:
1/2 = 1/4(1) - 1/3(1) + 1/2 + c/1
1/2 = 1/4 - 1/3 + 1/2 + c
By reducing the fractions to the same denominator and simplifying:
6 = 3 -4 + 6 + 12c
0 = -1 + 12c
12c = 1 c = 1/12
Let's subsititute c in the expression of y:
y = 1/4t² - 1/3t + 1/2 + 1/12/t²

Practice

Solve the differential equation dv/dt = 9.8 - 0.196v satisfying the condittion v(0) = 48

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