Question

First-Order Linear EquationsSolve the differential equations in Exercises 1–14.(t-1)³ ds/dt + 4(t-1)²s = t+1, t \u003e1

Accepted Answer

Rewrite the given differential equation in the standard linear form. Start by dividing both sides of the equation by $$ (t-1)^3$  to $$isolate $$ \frac{ds}{dt} . $$This gives:

$$ \frac{ds}{dt} + \frac{4}{t-1} s = \frac{t+1}{(t-1)^3$} $$ Identify the integrating factor $$ \mu(t) $$ for the linear differential equation. Recall that the integrating factor is given by:

$$ \mu(t) = e$$^{\int P(t) dt}$$ $$

where $$ P(t)  is $$the coefficient of $$ s  in $$the standard form. Here, $$ P(t) = \frac{4}{t-1} . $$Calculate the integrating factor:

$$ \mu(t) = e$$^{\int \frac{4}$${t-1} dt} = e$$^{4 \ln|t-1|}$$ = |t-1|^4 . $$Since $$ t \u003e 1 $$, we can write $$ \mu(t) = (t-1)^4$ . $$Multiply the entire differential equation by the integrating factor $$ (t-1)^4$  to $$make the left side an exact derivative:

$$ (t-1)^4$ \frac{ds}{dt} + 4 (t-1)^3$ s = (t-1)^4$ \cdot \frac{t+1}{(t-1)^3$} $$

which simplifies to:

$$ \frac{d}{dt} \left[ (t-1)^4$ s \right] = (t-1)(t+1) $$ Integrate both sides with respect to $$ t $$:

$$ \int \frac{d}{dt} \left[ (t-1)^4$ s \right] dt = \int (t-1)(t+1) dt $$

This will give you an expression for $$ (t-1)^4$ s $$ plus a constant of integration. Finally, solve for $$ s(t)  by $$dividing both sides by $$ (t-1)^4$ .$$

First-Order Linear Equations
Solve the differential equations in Exercises 1–14.

(t-1)³ ds/dt + 4(t-1)²s = t+1, t >1

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Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Domain and Continuity Conditions

First-Order Linear EquationsSolve the differential equations in Exercises 1–14.(t-1)³ ds/dt + 4(t-1)²s = t+1, t >1