Question

45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes

p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).

Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.

y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 4

Accepted Answer

Identify the given first-order linear differential equation: $$y'(t) + \frac{2t}{t^2$ + 1} y(t) = 1 + 3t^2$, with the initial condition y(1) = 4$. Determine the integrating factor p(t$$)$$ using the formula p(t$$) = \exp\left(\int a(t) \, dt\right)$$, where a(t$$) = \frac{2t}{$$t^2 + 1$$}$$. So, compute p(t$$) = \exp\left(\int \frac{2t}{$$t^2 + 1$$} \, dt\right)$$. Evaluate the integral inside the exponent to find p(t$$)$$. Notice that the integral involves a rational function that can be simplified by substitution. Multiply both sides of the original differential equation by the integrating factor p(t$$)$$ to rewrite the left side as the derivative of the product p(t) y(t$$)$$: $$\frac{d}{dt} \left(p(t) y(t)\right) = p(t) (1 + $$3t^2$$)$$. Integrate both sides with respect to t$ to get p(t) y(t$$) = \int p(t) (1 + $$3t^2$$) \, dt + C$$. Then solve for y(t$$)$$ by dividing both sides by p(t$$)$$, and use the initial condition y(1) = 4$ to find the constant C$.

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Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Initial Value Problems (IVP)