Question

Solve the differential equation. $$ f^{\prime \prime}(x)=x^{2}, f^{\prime}(0)=6, f(0)=3 $$

Short answer

The solution to the given differential equation is \(f(x) = \frac{1}{12}x^{4} + 6x + 3\)

Step-by-step solution

First Integration

We have \(f^{\prime \prime}(x) = x^{2}\). The first integration gives us \(f^{\prime}(x) = \frac{1}{3}x^{3} + C_1\), where \(C_1\) is the constant of the first integration.

Apply Initial Condition 1

Having obtained \(f^{\prime}(x)\), we can apply the first initial condition \(f^{\prime}(0) = 6\). By substituting these values into \(f^{\prime}(x) = \frac{1}{3}x^{3} + C_1\), we get \(C_1 = 6\). So, \(f^{\prime}(x) = \frac{1}{3}x^{3} + 6\).

Second Integration

Now we will integrate \(f^{\prime}(x) = \frac{1}{3}x^{3} + 6\) to find \(f(x)\). The second integration gives us \(f(x) = \frac{1}{12}x^{4} + 6x + C_2\), where \(C_2\) is the constant of the second integration.

Apply Initial Condition 2

Having obtained \(f(x)\), we can now apply the second initial condition \(f(0) = 3\). Substituting these values into \(f(x) = \frac{1}{12}x^{4} + 6x + C_2\), we obtain \(C_2 = 3\). Therefore, the solution to the given differential equation is \(f(x) = \frac{1}{12}x^{4} + 6x + 3\).