Please help me through diff equation questions

[shaneyj]

Can someone help me to get started...

Problem:
dy/dt + 2y(t) = 2x(t) where x(t) = u(t) and y(0) = -1

I know the standard form of a diff eq is:
dy/dx+ P(x)y=Q(x)

and I know how to solve for y using the integrating factor if the equation is in the same model as the standard equation

The issue I am having in the given problem: it is not in the form of the standard equation.
I don't see that 2y(t) is = to P(x)y.

And beyond that, I am unclear on where to plug in the givens

Any help would be appreciated.

I've attached a PDF if that helps

 

[Ratch]

Can someone help me to get started...

Problem:
dy/dt + 2y(t) = 2x(t) where x(t) = u(t) and y(0) = -1

I know the standard form of a diff eq is:
dy/dx+ P(x)y=Q(x)

and I know how to solve for y using the integrating factor if the equation is in the same model as the standard equation

The issue I am having in the given problem: it is not in the form of the standard equation.
I don't see that 2y(t) is = to P(x)y.

And beyond that, I am unclear on where to plug in the givens

Any help would be appreciated.

I've attached a PDF if that helps

OK, x(t) = u(t) , which is the unit step function. For t > 0, you have to solve the DE dy/dt + 2y(t) = 2 with the given condition. For t < 0, you have to solve the DE dy/dt + 2y(t) = 0 with the given condition. Looks like the separation of variables method will work. Ask if you have any more probs.

Ratch

 

[shaneyj]

OK, x(t) = u(t) , which is the unit step function. For t > 0, you have to solve the DE dy/dt + 2y(t) = 2 with the given condition. For t < 0, you have to solve the DE dy/dt + 2y(t) = 0 with the given condition. Looks like the separation of variables method will work. Ask if you have any more probs.

Ratch

Thank you, Ratch.
Why is it that for t greater than zero the DE is equal to 2?

 

[Ratch]

Thank you, Ratch.
Why is it that for t greater than zero the DE is equal to 2?

The unit step function u(t) is 0 when t < 0 and 1 when t > 0 . Therefore, if x(t) = u(t), 2 x(t) = 2 when t > 0 and 0 when t< 0. So you have to solve the DE twice for each range of t.

Ratch

 

[shaneyj]

The unit step function u(t) is 0 when t < 0 and 1 when t > 0 . Therefore, if x(t) = u(t), 2 x(t) = 2 when t > 0 and 0 when t< 0. So you have to solve the DE twice for each range of t.

Ratch

I get that...

I solved and get y= 1/2 + C/2e^2t

So for y(0)=-1, should I set the equation equal to -1, plug in 0 for the t value, and find C?
Then set the equation = 0 and 2, like you originally indicated, and find t for those?
Am I correct in thinking that once I find C that value will be constant for all values of y, or does it change?

 

[Ratch]

I get that...

I solved and get y= 1/2 + C/2e^2t

So for y(0)=-1, should I set the equation equal to -1, plug in 0 for the t value, and find C?
Then set the equation = 0 and 2, like you originally indicated, and find t for those?
Am I correct in thinking that once I find C that value will be constant for all values of y, or does it change?

You are going to have to learn how to do this, or you won't get very far in DE.

Ratch

 

[shaneyj]

You are going to have to learn how to do this, or you won't get very far in DE.

Ratch

View attachment 36621

That's great. Thank you. I'm going to work it with that method.
Our text says to use the method where e raised to the integral of P(x) (the integrating factor) is multiplied on both sides of the equation which is supposed to leave the IF times the y term equal to the other side times the IF.
That's where I got my answer.
It's been almost 2 years since I took calculus. I do need a refresher.

 

[Ratch]

That's great. Thank you. I'm going to work it with that method.
Our text says to use the method where e raised to the integral of P(x) (the integrating factor) is multiplied on both sides of the equation which is supposed to leave the IF times the y term equal to the other side times the IF.
That's where I got my answer.
It's been almost 2 years since I took calculus. I do need a refresher.

Here is the solution using an integrating factor. You can't fool Mother Math, the solution will be the same.

Ratch