Question

1. (exam 4 review 5b) solve the initial value problem: $g(t)=\frac{2}{t^{2}}$; $g(-1)=g(1)=1$.

Explanation:

Step1: Integrate \(g''(t)\) to find \(g'(t)\)

Integrate \(g''(t)=\frac{2}{t^{2}} = 2t^{-2}\) with respect to \(t\).
Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(g'(t)=\int2t^{-2}dt=2\frac{t^{-2 + 1}}{-2 + 1}+C_1\).
So, \(g'(t)=-\frac{2}{t}+C_1\).

Step2: Use the initial condition \(g'(-1)=1\) to find \(C_1\)

Substitute \(t=-1\) and \(g'(-1) = 1\) into \(g'(t)=-\frac{2}{t}+C_1\).
\(1=-\frac{2}{-1}+C_1\), which simplifies to \(1 = 2+C_1\).
Solving for \(C_1\), we get \(C_1=-1\). So, \(g'(t)=-\frac{2}{t}-1\).

Step3: Integrate \(g'(t)\) to find \(g(t)\)

Integrate \(g'(t)=-\frac{2}{t}-1\) with respect to \(t\).
Using \(\int\frac{1}{x}dx=\ln|x|+C\) and \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(g(t)=\int(-\frac{2}{t}-1)dt=-2\ln|t|-t+C_2\).

Step4: Use the initial condition \(g(1) = 1\) to find \(C_2\)

Substitute \(t = 1\) and \(g(1)=1\) into \(g(t)=-2\ln|t|-t+C_2\).
\(1=-2\ln(1)-1+C_2\). Since \(\ln(1)=0\), we have \(1=- 0-1+C_2\).
Solving for \(C_2\), we get \(C_2=2\).

Answer:

\(g(t)=-2\ln|t|-t + 2\)