Antiderivatives and Initial Value Problems
Solved Problems
Example 7.
Solve the initial value problem \[\frac{{dy}}{{dx}} = \frac{2}{{x + 1}}, y\left( 0 \right) = 2.\]
Solution.
Find the general antiderivative of \(\frac{2}{{x + 1}}:\)
Use the initial condition \(y\left( 0 \right) = 2\) to evaluate the constant \(C:\)
So the particular solution is written as
Example 8.
Solve the initial value problem \[\frac{{dr}}{{d\theta}} = \cos \frac{\theta }{2}, r\left( {\frac{\pi }{3}} \right) = 2.\]
Solution.
We first write the general antiderivative of the function \(\cos \frac{\theta }{2}:\)
Substitute the initial condition \(r\left( {\frac{\pi }{3}} \right) = 2\) to find the constant \(C:\)
Hence, the particular solution of the differential equation is given by
Example 9.
Solve the initial value problem \[\frac{{dz}}{{dt}} = \cos t - 2\sin t, z\left( 0 \right) = 5.\]
Solution.
The general antiderivative of the function \(\cos t - 2\sin t\) is
Substitute the initial condition \(z\left( 0 \right) = 5\) to evaluate the constant \(C:\)
So the solution of the initial value problem is given by
Example 10.
A function \(y\left( x \right)\) is given by the differential equation \[\frac{{dy}}{{dx}} = \frac{1}{x} + 2x\] with the initial condition \(y\left( 1 \right) = 0.\) Find the function value when \(x = e.\)
Solution.
Write the general antiderivative of the function \(\frac{1}{x} + 2x:\)
Using the initial condition \(y\left( 1 \right) = 0,\) determine the constant \(C:\)
Then the function \(y\left( x \right)\) is given by
Compute the function value at \(x = e:\)
Example 11.
A ball dropped from the top of a tall building has velocity \[v\left( t \right) = -10t - 5 \, \frac{\text{m}}{\text{s}}.\] Find the height of the building given that the ball strikes the ground after \(t = 4\) seconds.
Solution.
Let \(H\) be the height of the building. To get the vertical position of the ball \(h\left( t \right),\) we determine the general antiderivative of the velocity function:
The constant \(C\) is found from the initial condition \(h\left( 0 \right) = H.\) Hence
After \(t = 4\) seconds, the height \(h\) is equal to \(0,\) so we have