How Can I Solve the Differential Equation for Y in This Formula?

[yamdizzle]

I have

(θ$_{1}$-θ$_{2}$)*exp(r*t) + r* Y = dY/dt

How can I find Y?
I tried to reverse the f ' g +g' f but I keep getting an extra term

Thanks

 

[LCKurtz]

I have

(θ$_{1}$-θ$_{2}$)*exp(r*t) + r* Y = dY/dt

How can I find Y?
I tried to reverse the f ' g +g' f but I keep getting an extra term

Thanks

Write it as
$$y' - ry = (\theta_1-\theta_2)e^{rt}$$ Multiply both sides by e^-rt and you should have an exact derivative on the left side.

 

[yamdizzle]

Couldn't quite get it? What do you mean by exact derivative?

so what is y?

 

[LCKurtz]

Couldn't quite get it? What do you mean by exact derivative?

so what is y?

Show us what happened when you followed my advice. The left side should look like the derivative of a product. What did you get?

 

[yamdizzle]

so we have:

exp(-r*t) y' - exp(-r*t) r y = theta1 - theta2

I assume you mean
f = exp(-r*t)
g = y

but I'm not sure how to place thetas

so I think y will have a exp(r*t) in it but not sure about the rest.

 

[LCKurtz]

so we have:

exp(-r*t) y' - exp(-r*t) r y = theta1 - theta2

I assume you mean
f = exp(-r*t)
g = y

but I'm not sure how to place thetas

so I think y will have a exp(r*t) in it but not sure about the rest.

Yes, so the left side is (e^-rty)' so your equation is: $$(e^{-rt}y)'=\theta_1-\theta_2$$ I assume the thetas are constant. So what do you get when you integrate both sides with respect to t and solve for y?

 

[yamdizzle]

Yep, got it.

Thanks