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For the following problems, determine how parameter $a$ affects the solution.Solve $y^{\prime}-y=e^{k t}$ with the initial condition $y(0)=0 .$ As $k$ approaches $1,$ what happens to your formula?

$y(t)=\frac{1}{k-1}\left(e^{k t}-e^{t}\right)$$\lim _{k \rightarrow 1} y(t)=e^{t}$

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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Section 45 problem to 61 s. So what you see here is that we have, ah, first order linear, ordinary differential equation and were asked to solve this. The key to having this is in standard form, which this one actually is in standard form. So the integrating factor has a function of t is going to be e and then so minus one DT. So this is you to the minus T is my integrating factor. So this equation transforms into why e to the minus t prime is equal to e to the k t you to the minus t. So this is you to the t times K minus one. Now, in order to solve this equation, I will integrate both sides of this equation on the left side, I just get y e to the minus. T is equal to and then on the right side. When you integrate the issue, get E to the K minus one T and then times one over K minus one, plus a constant of integration. Therefore, why is equal to one over K minus one you to the K minus one e e to the T plus C to the T. And so this just becomes, um hi is equal to one over K minus one e to the k t plus c e to the t. Now. We had the boundary condition that why of zero, um, is equal to zero. That would allow me to solve for the constant of integration. So why to the zero is zero. So that means that zero is equal to one over, OK, minus one. And then e to the zero is just one plus ce to the zero on. It's just simply one. And so what you see here is that c You've got a minus one over K minus one. So this solution becomes why the could one over K minus one you to the K T minus one, over K minus one. You to the t factor out that one over K minus one. So one overcame IRS one e to the K T minus e to the T. Now we're asked to take a look and see what would happen as I let kay um, approach zero. Okay, so what would be the limit as K approaches zero in this case, Um, excuse me as can purchase one. So when K approaches one, you know that the denominator Here's if I look at this, the limit is K approaches one of e to the K T minus e to the t over K minus one. So as K approaches one, I'm approaching zero and the denominator and I've had e to the team and is either the t I'm approaching zero in the numerator. So it's the indeterminant form. So using low petals rule, I can differentiate and then take the derivative. So this is also going to be equal to the limit has que approaches one and we have to differentiate with respect to K. So differentiating with respect to K E to the K T. That's going to be easy to the K t. And if K is my variable, that's going to give me a t e to the K T minus. When you differentiate eat of the tea with respect to K that zero and then in the denominator when you different ship with respect to K, that is one. So this is the limit as que approaches zero of t e to the k t. Excuse me que approaches one I'm just so accustomed to zero as K approaches one. So when K approaches one, you can simply substitute that in. Now, this is just t e to the t. So that's what the function turns into as K approaches one in this case.

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