Solve a Differential Equation

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January 11th 2010, 08:28 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.
January 11th 2010, 08:32 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.

$\frac{dy}{dt}=-ay+b=-a\left(\frac{-ay}{-a}+\frac{b}{-a}\right)=-a\left(y-\frac{b}{a}\right)$ cross multiplying gives $\frac{dy}{y-\frac{b}{a}}=-a\text{ }dt$ . Integrate both sides.
January 11th 2010, 08:36 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

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

Can be written as

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

taking out -a as a factor

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

Then separating

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

Now integrate both sides
January 11th 2010, 09:26 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?
January 12th 2010, 05:52 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.
January 12th 2010, 06:55 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.