How to Solve an Initial-Value Problem?

[ssb]

Homework Statement

Solve the following initial-value problem
$$\(y'+5y=1,\:y(1)=0\)$$

The Attempt at a Solution

y' + 5y = 0 (this is to solve for a general solution)

y = -(5/2)y^2
-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_

y' + 5y = 1 (this is for the particular solution)

y = 1/5

Therefore when you add the 2 equations up the answer should be

1/5 - (5/2)y^-2

But I went wrong somewhere.
My math tutor said that the answer involves e somehow but frankly I am kinda lost here. Thanks.

 

[Päällikkö]

y' = dy/dt
So if you go with y = -(5/2)y^2, y' = 0. This would mean that the original differential equation isn't satisfied.

What methods do you have at your disposal? Are you familiar with separation of variables?
Write the equation as
y' + 5y = 1
dy/dt + 5y = 1
dy/(1-5y) = dt
Does that ring any bells?

EDIT: You could use x instead of t if you're more used to that.

 

[HallsofIvy]

Homework Statement

Solve the following initial-value problem
$$\(y'+5y=1,\:y(1)=0\)$$

The Attempt at a Solution

y' + 5y = 0 (this is to solve for a general solution)

y = -(5/2)y^2
-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_

Olay, how in the world did you get this?
y is a function of x not y!
If you meant y= - (5/2)x², that obviously doesn't work:
If y= -(5/2)x², then y'= -5x, not -5 y.
To solve y'= -5y write it as dy/y= -5dx and integrate both sides.

y' + 5y = 1 (this is for the particular solution)

y = 1/5

Therefore when you add the 2 equations up the answer should be

1/5 - (5/2)y^-2

But I went wrong somewhere.
My math tutor said that the answer involves e somehow but frankly I am kinda lost here. Thanks.

 

[HallsofIvy]

You got this in before looking at my respons. You want y as a function of x, not y!

1. Integrate dy/y= -5 dx

2. Sollve for y.

 

[EugP]

I think the easiest way to do it is like Päällikkö said, separation of variables.
Just do the following:

$$y' + 5y = 1$$

$$y' = 1 -5y$$

$$\frac{dy}{1 - 5y} = 1 dt$$

$$\int\frac{dy}{1 - 5y} = \int1 dt$$

$$-\frac{1}{5}\ln(1 - 5y) = t + C$$

$$\ln(1 - 5y) = -5t + C$$

$$1 -5y = Ce^{-5t}$$

$$5y = Ce^{-5t} - 1$$

$$y = \frac{1 - Ce^{-5t}}{5}$$

Now use the initial condition to find C:

$$0 = \frac{1 - Ce^{-5}}{5}$$

$$0 = 1 - Ce^{-5}$$

$$1 = Ce^{-5}$$

$$\frac{1}{e^{-5}} = C$$

$$C = e^{5}$$

Now put it all together:

$$y = \frac{1 - e^{5}e^{-5t}}{5}$$

Simplify and you get:

$$y = \frac{1 - e^{5 (1 - t)}}{5}$$