Initial Value Differential Equations problem

[Chandasouk]

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) = Ce^t

y-5/2 = Ce^t

solving for C

(y₀) - 5/2=C*e⁰

C= y₀ - 5/2

 

[Mark44]

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)e^2t.

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 = e^{2t + C} = Ae^2t, where A = e^C

==> y = 5/2 + Ae^2t

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) = Ce^t


y-5/2 = Ce^t

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.

 

[Chandasouk]

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\int du/u = \intdt

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

 

[Mark44]

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\int du/u = \intdt

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

e both sides

1/2(2y-5) = Cet

You're making a mistake here (above). e^u*v \neq e^u * u^v.

Also, e^1/2 \neq 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

 

[Chandasouk]

So when I e both sides of 1/2*ln|2y - 5| = t + C

it's should be

e^1/2*e^{ln|2y - 5|} = e ^{t + C}

e^1/2*2y-5=Ce^t

2y-5 = Ce^t/e^1/2

2y-5 = Ce^2t

Now I see where the 2t came from.

 

[Mark44]

So when I e both sides of 1/2*ln|2y - 5| = t + C

it's should be

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

which is what you're trying to do.

e^1/2*2y-5=Ce^t

2y-5 = Ce^t/e^1/2

2y-5 = Ce^2t

Now I see where the 2t came from.

No you don't.
\frac{e^t}{e^{1/2}} \neq e^{2t}