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Differential Equations : Initial-Value Problems

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Example Questions

Example Question #11 : Initial Value Problems

Solve the following initial value problem: y' = 10(y - t) , y(0) = 1 .

Possible Answers:

$y = t + \frac{1}{10}$

$y = t + 10e^{10t}$

$y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$

$y = t + \frac{9e^{10t}}{10}$

$y = t + \frac{1}{10} + 10e^{10t}$

Correct answer:

$y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$

Explanation:

This type of problem will require an integrating factor. First, let's get it into a better form: y' - 10y = -10t . Once we have an equation in the form y' + p(t)y = q(t) , we find $\mu = e^{\int p dt} = e^{\int -10 dt} = e^{-10t}$ (where the constant of integration is omitted because we only need one, arbitrary integrating factor).

Once we do this, we can see that $(\mu y)' = \mu y' + \mu'y = e^{-10t}y' - 10e^{-10t}y = e^{-10t}(y' - 10t) = e^{-10t}\cdot-10t$

This is simply due to product rule, and then at the end, substitution of the original equation. Thus, as we know that $(\mu y)' = \mu q$ , we can just integrate both sides to find y.

$e^{-10t}y = \int-10te^{-10t}dt$ . A quick application of integration by parts with u = -10t and $dv = e^{-10t}$ tells us that the right hand side is $te^{-10t} + \frac{e^{-10t}}{10}+C$ . Dividing both sides by mu, we are left with $y = t + \frac{1}{10} + \frac{C}{e^{-10t}}$ . Plugging in the initial condition, we find $1 = \frac{1}{10} + C$ giving us $C = \frac{9}{10}$ . Thus, we have a final answer of $y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$ .

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Differential Equations : Initial-Value Problems

Study concepts, example questions & explanations for Differential Equations

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Example Questions

Example Question #11 : Initial Value Problems

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