Question

solve the differential equation.
$(t - 1)^{7}s + 8(t - 1)^{6}s = t + 1,t>1$
the solution is $s = \square$.

Explanation:

Step1: Rewrite the differential equation

Divide the given equation \((t - 1)^{7}s'+8(t - 1)^{6}s=t + 1\) by \((t - 1)^{7}\) to get it in the form \(s'+p(t)s = q(t)\).
We have \(s'+\frac{8}{t - 1}s=\frac{t + 1}{(t - 1)^{7}}\).

Step2: Find the integrating factor

The integrating factor \(\mu(t)=e^{\int p(t)dt}\). Here \(p(t)=\frac{8}{t - 1}\), so \(\int\frac{8}{t - 1}dt=8\ln(t - 1)=\ln((t - 1)^{8})\) (since \(t>1\)). Then \(\mu(t)=e^{\ln((t - 1)^{8})}=(t - 1)^{8}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(s'+\frac{8}{t - 1}s=\frac{t + 1}{(t - 1)^{7}}\) by \((t - 1)^{8}\).
Using the product rule \((uv)'=u'v+uv'\), where \(u = s\) and \(v=(t - 1)^{8}\), we get \(\frac{d}{dt}(s(t - 1)^{8})=(t + 1)(t - 1)\).
Since \((t + 1)(t - 1)=t^{2}-1\).

Step4: Integrate both sides

Integrate \(\frac{d}{dt}(s(t - 1)^{8})=t^{2}-1\) with respect to \(t\).
\(\int\frac{d}{dt}(s(t - 1)^{8})dt=\int(t^{2}-1)dt\).
We know that \(\int(t^{2}-1)dt=\frac{t^{3}}{3}-t+C\).
So \(s(t - 1)^{8}=\frac{t^{3}}{3}-t+C\).

Step5: Solve for \(s\)

Divide both sides by \((t - 1)^{8}\) to get \(s=\frac{t^{3}-3t + 3C}{3(t - 1)^{8}}\).

Answer:

\(s=\frac{t^{3}-3t + C}{3(t - 1)^{8}}\)