Question

In Exercises \(11-20,\) solve the initial value problem explicitly. \(\frac{d x}{d t}=\frac{1}{t}-\frac{1}{t^{2}}+6\) and \(x=0\) when \(t=1\)

Short answer

The particular solution to the given initial value problem is \( x(t) = \ln|t| + \frac{1}{t} + 6t - 7 \)

Step-by-step solution

Integrate the differential equation

Start by integrating both sides of the given differential equation with respect to \( t \), \(\int \frac{d x}{d t} dt = \int (\frac{1}{t}-\frac{1}{t^{2}}+6) dt\). After integrating, we get \(x(t) = \ln|t| + \frac{1}{t} + 6t + C\), where \(C\) is the constant of integration.

Substitute the initial value

To determine the value of the constant \(C\), we substitute the given initial condition \(x = 0\) when \(t = 1\). So, we substitute these values into our integrated equation \(0 = \ln|1| + \frac{1}{1} + 6(1) + C\), which when simplify we get \(C = -7\).

Write down the final solution

Substitute \(C = -7\) into the general solution to get the particular solution of the initial value problem. This gives the solution: \( x(t) = \ln|t| + \frac{1}{t} + 6t - 7 \).