- #1

asdf1

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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...

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- Thread starter asdf1
- Start date

- #1

asdf1

- 734

- 0

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...

- #2

Emieno

- 96

- 0

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.

- #3

Galileo

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- #4

Emieno

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Galileo, are you really well today ?

- #5

cronxeh

Gold Member

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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

- #6

Galileo

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

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.

- #7

asdf1

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:P

sorry my mistake~

dy/dt +1/2y=1/2e^(t/3)

sorry my mistake~

dy/dt +1/2y=1/2e^(t/3)

- #8

HallsofIvy

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In this case, the equation is

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

That tells us that if we multiply the whole equation by e

e

The left hand side is just d(e

Integrating both sides, e

y(t)= (6/5)e

Notice the e

- #9

mantito

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- #10

HallsofIvy

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Your're right, I did. Thanks.

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