# Differential equation problem (1 Viewer)

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#### asdf1

there's an example in my textbook,
solve the differential equation
dy/dx + 1/2y =1/2e^(t/3)

this problem is weird
because i can't separate the variables on both sides and integrate them...

#### Emieno

Did you have any careful check again your book or possibly think of something that might be incorrect on the right hand side of your equation anyway ?
If :x: is in place of :t:, it is easily sovable and otherwise, it is going to be different since you have to explain how things are to logically work out in the presence of t.

#### Galileo

Science Advisor
Homework Helper
There are other ways to get the solution. The nice thing about exponentials is that its derivative is also an exponential. Try a function of the form: y(t)=Aexp(t/3) to find a particular solution.

#### Emieno

Galileo, are you really well today ????????

#### cronxeh

Gold Member
dy/dx + 1/2y =1/2e^(t/3)

Are you sure this is the problem??

Usually for ODE the equation takes this form:
http://www.ucl.ac.uk/Mathematics/geomath/level2/deqn/int2.gif

So you right hand side should be a function of x, not t. Make sure you re-read your problem because otherwise this is not an ODE

#### Galileo

Science Advisor
Homework Helper
Emieno said:
Galileo, are you really well today ????????
I'm fine thanks.

I assumed the dy/dx is a typo. The equation might have been y'+1/2y=exp(t/3) and by force of habit asdf1 typed dy/dx instead of dy/dt.

Anyway, assuming a solution of the form Aexp(t/3) will get you a particular solution. Add to that a solution of y'+1/2y=0 for the general one.

#### asdf1

:P
sorry my mistake~
dy/dt +1/2y=1/2e^(t/3)

#### HallsofIvy

Science Advisor
Okay, assuming you mean dy/dt+ (1/2)y= (1/2)e(t/3), that's a linear equation. Not every equation is "separable". If you are doing first order linear equations you should already have learned about exact equations and "integrating factors".

In this case, the equation is not exact- that is, there is no function f(y) so that d(f(y))/dt= dy/dt+ (1/2)y, but we can find a function u(t) so that, after we multiply the entire equation by u, it is exact. We want to find u(t) so that
d(u(t)y)/dt= u(t)(dy/dt)+ (1/2)u(t)y. Applying the product rule to the left hand side of that, d(u(t)y)/dt= u(t)(dy/dt)+ (du/dt)y= u(t)(dy/dt)+ (1/2)u(t)y.
That means we must have du/dt= (1/2)u which is separable: In fact it is easy to see that u(t)= e(1/2)t.

That tells us that if we multiply the whole equation by e(1/2)t, we get
e(1/2)t(dy/dt)+ (1/2)e(1/2)ty= e(1/2+ (1/3))t
The left hand side is just d(e(1/2)ty)/dt and the right hand side is e(5/6)t. In other words, d(e(1/2)ty)/dt= e(5/6)t.

Integrating both sides, e(1/2)ty= (6/5)e(5/6)t+ C so

y(t)= (6/5)e(1/3)t+ Ce(-1/2)t.

Notice the e(1/3)t in there, exactly as asdf1 suggested.

#### mantito

you forgot $$\frac{1}{2}$$ in front of $$e^\frac{t}{3}$$. The answer should be $$y=\frac{3}{5}e^\frac{t}{3}+Ce^-\frac{t}{2}$$

#### HallsofIvy

Science Advisor
Your're right, I did. Thanks.

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