# Thread: Seemingly easy differential equation, but I got confused

1. ## Seemingly easy differential equation, but I got confused

Solve:

2 (dy/dt)=2y+3t

So I divide through by 2:

(dy/dt)=y+(3/2)t

I take the homogenous yh=e^t

Then I make it

(dy/dt)-y=(3/2)t

So I guess a*t

Plug that in to get:

a-at=(3/2)t
So that

a(1-t)=(3/2)t

and then

a=(3/2)t/(1-t)

And multiplying that my t and combing with the homogeous I get:

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

but I've never seen an answer like this since usually it is a more complicated function with es and cosines....

Is this correct or where did I make a mistake?

Thanks.

2. Originally Posted by zhupolongjoe
Solve:

2 (dy/dt)=2y+3t

So I divide through by 2:

(dy/dt)=y+(3/2)t

I take the homogenous yh=e^t

Then I make it

(dy/dt)-y=(3/2)t

So I guess a*t

Plug that in to get:

a-at=(3/2)t
So that

a(1-t)=(3/2)t

and then

a=(3/2)t/(1-t)

And multiplying that my t and combing with the homogeous I get:

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

but I've never seen an answer like this since usually it is a more complicated function with es and cosines....

Is this correct or where did I make a mistake?

Thanks.
First off, a is a constant so you can't solve for a as a function of t.

Let $\displaystyle y_p = at + b$ so

$\displaystyle a -at -b = \frac{1}{2}t$
so $\displaystyle a-b = 0, -a = \frac{1}{2}$ so $\displaystyle a = b = - \frac{1}{2}$ and $\displaystyle y_p = -\frac{1}{2} t - \frac{1}{2}$