# Solve a Differential Equation

• Jan 11th 2010, 07:28 PM
sfgiants13
Solve a Differential Equation
The equation is dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't seem to get the equation from the 1st form to the 2nd for some reason. I haven't taken calc in over a year so I'm a bit rusty.
• Jan 11th 2010, 07:32 PM
Drexel28
Quote:

Originally Posted by sfgiants13
The equation is dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't seem to get the equation from the 1st form to the 2nd for some reason. I haven't taken calc in over a year so I'm a bit rusty.

$\displaystyle \frac{dy}{dt}=-ay+b=-a\left(\frac{-ay}{-a}+\frac{b}{-a}\right)=-a\left(y-\frac{b}{a}\right)$ cross multiplying gives $\displaystyle \frac{dy}{y-\frac{b}{a}}=-a\text{ }dt$. Integrate both sides.
• Jan 11th 2010, 07:36 PM
pickslides
dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't

$\displaystyle \frac{dy}{dt} = -ay+b$

Can be written as

$\displaystyle \frac{dy}{dt} = -ay+\frac{-a}{-a}\times b$

taking out -a as a factor

$\displaystyle \frac{dy}{dt} = -a\left(y+\frac{b}{-a}\right)$

Then separating

$\displaystyle \frac{dy}{\left(y+\frac{b}{-a}\right)} = -a~dt$

Now integrate both sides
• Jan 11th 2010, 08:26 PM
sfgiants13
Thanks guys I get that now. Next part asks to sketch the solution for several different initial conditions. Is that just a matter of plugging in real numbers for a and b and sketching them?
• Jan 12th 2010, 04:52 AM
mr fantastic
Quote:

Originally Posted by sfgiants13
Thanks guys I get that now. Next part asks to sketch the solution for several different initial conditions. Is that just a matter of plugging in real numbers for a and b and sketching them?

No. As far as I can see, a and b have nothing to do with the initial conditions.

I hope you realise that when you integrate, an arbitrary constant of integration is introduced. The value of that constant can be determined if intitial conditions are given eg. y = 1 when t = 0.

By the way, if you're studying differential equations you better make it your urgent business to rust rid your calculus.
• Jan 12th 2010, 05:55 AM
Calculus26
Note for dy/dt =- ay + b

y = b/a is an equiilbrium solution.

your initial conditions do depend on a and b in the sense that if y(0) < b/a

then the solution curve approaches the line y= b/a from below.

If y(0) > b/a then the solution curve decays to the line y =b/a from above.