Calculus

Integration of Functions

Integration of Functions Logo

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 \(\frac{2}{{x + 1}}:\)

\[y = 2\ln \left| {x + 1} \right| + C.\]

Use the initial condition \(y\left( 0 \right) = 2\) to evaluate the constant \(C:\)

\[2\ln \left| {0 + 1} \right| + C = 2,\;\; \Rightarrow 2 \cdot \ln 1 + C = 2,\;\; \Rightarrow 2 \cdot 0 + C = 2,\;\; \Rightarrow C = 2.\]

So the particular solution is written as

\[y = 2\ln \left| {x + 1} \right| + 2.\]

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 \(\cos \frac{\theta }{2}:\)

\[r\left( \theta \right) = 2\sin \frac{\theta }{2} + C.\]

Substitute the initial condition \(r\left( {\frac{\pi }{3}} \right) = 2\) to find the constant \(C:\)

\[2\sin \frac{{\frac{\pi }{3}}}{2} + C = 2,\;\; \Rightarrow 2\sin \frac{\pi }{6} + C = 2,\;\; \Rightarrow 2 \cdot \frac{1}{2} + C = 2,\;\; \Rightarrow 1 + C = 2,\;\; \Rightarrow C = 1.\]

Hence, the particular solution of the differential equation is given by

\[r\left( \theta \right) = 2\sin \frac{\theta }{2} + 1.\]

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

\[z\left( t \right) = \sin t + 2\cos t + C.\]

Substitute the initial condition \(z\left( 0 \right) = 5\) to evaluate the constant \(C:\)

\[\sin 0 + 2\cos 0 + C = 5,\;\; \Rightarrow 0 + 2 \cdot 1 + C = 5,\;\; \Rightarrow C = 3.\]

So the solution of the initial value problem is given by

\[z\left( t \right) = \sin t + 2\cos t + 3.\]

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 \(\frac{1}{x} + 2x:\)

\[y = \ln x + {x^2} + C.\]

Using the initial condition \(y\left( 1 \right) = 0,\) determine the constant \(C:\)

\[y\left( 1 \right) = 0,\;\; \Rightarrow \ln 1 + {1^2} + C = 0,\;\; \Rightarrow 0 + 1 + C = 0,\;\; \Rightarrow C = - 1.\]

Then the function \(y\left( x \right)\) is given by

\[y = \ln x + {x^2} - 1.\]

Compute the function value at \(x = e:\)

\[y\left( e \right) = \ln e + {e^2} - 1 = \cancel{1} + {e^2} - \cancel{1} = {e^2}.\]

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:

\[h\left( t \right) = C - 5{t^2} - 5t.\]

The constant \(C\) is found from the initial condition \(h\left( 0 \right) = H.\) Hence

\[h\left( t \right) = H - 5{t^2} - 5t.\]

After \(t = 4\) seconds, the height \(h\) is equal to \(0,\) so we have

\[H - 5 \cdot {4^2} - 5 \cdot 4 = 0,\;\; \Rightarrow H = 80 + 20 = 100\,\text{m}.\]
Page 1 Page 2