• Support PF! Buy your school textbooks, materials and every day products Here!

Initial Value Differential Equations problem

  • Thread starter Chandasouk
  • Start date
  • #1
165
0
dy/dt = 2y-5, y(0) = y_0

The answer I got for this was y =5/2 + (y_0-5/2)e^t

but the solution is y =5/2 + (y_0-5/2)e^-2t

Where did this -2t come from?

My steps:

dy = (2y-5)dt Integrating both sides gives

1/2*ln|2y - 5| = t + C take exp of both sides gives

1/2(2y-5) = Cet

y-5/2 = Cet

solving for C

(y0) - 5/2=C*e0

C= y0 - 5/2

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
33,514
5,195
dy/dt = 2y-5, y(0) = y_0

The answer I got for this was y =5/2 + (y_0-5/2)e^t

but the solution is y =5/2 + (y_0-5/2)e^-2t
Both answers above are incorrect. Did you write the book's solution incorrectly? The correct solution isd be y = 5/2 + (y0 - 5/2)e2t.
Where did this -2t come from?

My steps:

dy = (2y-5)dt Integrating both sides gives
You're skipping some steps here. Here's what I did.
dy/(2y - 5) = dt
==> (1/2)dy/(y - 5/2) = dt (factoring 2 out of the denominator on the left)
==> dy/(y - 5/2) = 2dt
Now integrate, getting
ln(y - 5/2) = 2t + C
==> y - 5/2 = e2t + C = Ae2t, where A = eC

==> y = 5/2 + Ae2t

All that's left is to use the initial condition to solve for A.

1/2*ln|2y - 5| = t + C take exp of both sides gives

1/2(2y-5) = Cet


y-5/2 = Cet

solving for C

(y0) - 5/2=C*e0

C= y0 - 5/2
Whenever you solve a differential equation, you should verify that your solution actually is a solution. If you had done that, you would have seen that neither the solution you found nor the one you said was "the solution" actually works.
 
  • #3
165
0
I have misread the solution. The answer is y =5/2 + (y_0-5/2)e^2t

I would've never thought to factor out the 2 like you did though. I just integrated as

dy/(2y-5) = dt

Used U substitution for the integral on the left side of the equation.

U = 2y-5, du = 2dy, so du/2=dy

that gave me 1/2[tex]\int[/tex] du/u = [tex]\int[/tex]dt

1/2*ln|2y - 5| = t + C

e both sides

1/2(2y-5) = Cet

y-5/2 = Cet

solving for C

(y0) - 5/2=C*e0

C= y0 - 5/2

However I cannot get the 2t to appear as my exponent
 
  • #4
33,514
5,195
I have misread the solution. The answer is y =5/2 + (y_0-5/2)e^2t

I would've never thought to factor out the 2 like you did though. I just integrated as

dy/(2y-5) = dt

Used U substitution for the integral on the left side of the equation.

U = 2y-5, du = 2dy, so du/2=dy

that gave me 1/2[tex]\int[/tex] du/u = [tex]\int[/tex]dt

1/2*ln|2y - 5| = t + C

e both sides

1/2(2y-5) = Cet
You're making a mistake here (above). eu*v [itex]\neq[/itex] eu * uv.

Also, e1/2 [itex]\neq[/itex] 1/2.
y-5/2 = Cet

solving for C

(y0) - 5/2=C*e0

C= y0 - 5/2

However I cannot get the 2t to appear as my exponent
 
  • #5
165
0
So when I e both sides of 1/2*ln|2y - 5| = t + C

it's should be

e1/2*eln|2y - 5| = e t + C

e1/2*2y-5=Cet

2y-5 = Cet/e1/2

2y-5 = Ce2t

Now I see where the 2t came from.
 
  • #6
33,514
5,195
So when I e both sides of 1/2*ln|2y - 5| = t + C

it's should be

e1/2*eln|2y - 5| = e t + C
NO! That's what I said in my previous post.

Starting with 1/2*ln|2y - 5| = t + C, the simplest thing to do is to multiply both sides by 2.

But
[tex]e^{1/2 * ln|2y - 5|} \neq e^{1/2}~e^{ ln|2y - 5|}[/tex]

which is what you're trying to do.
e1/2*2y-5=Cet

2y-5 = Cet/e1/2

2y-5 = Ce2t

Now I see where the 2t came from.
No you don't.
[tex]\frac{e^t}{e^{1/2}} \neq e^{2t}[/tex]
 

Related Threads on Initial Value Differential Equations problem

Replies
3
Views
778
Replies
2
Views
1K
Replies
15
Views
555
Replies
2
Views
1K
Replies
11
Views
2K
Replies
4
Views
893
Replies
14
Views
2K
Replies
5
Views
2K
Replies
9
Views
2K
Replies
1
Views
666
Top