Understanding Separable Differential Equations and Initial Value Problems

[simba_]

Homework Statement

I am stuck on these two questions. The first one I can start off and finish but i cannot do the middle part and in the second question I have no idea how to start it off.

Find the general solution of the following separable equations; then use the solution which obeys the initial condition y(1)=2 to solve

1) tyy'=y-1

2) y'=y²-4y

Homework Equations

The Attempt at a Solution

1)

yy'/y-1 = 1/t
y/(y-1) dy = 1/t dt

here i do not know how to integrate the LHS??
but i looked the answer up in wolfram alpha and got it solved from there.

2) no idea how to start this one

the answer for question is
-1/(c+logt)

-where c is 1/2

Sorry for my poorly presented questions, Ill learn how to write out the equations properly for my next post!

 

[fzero]

For

y/(y-1) dy = 1/t dt

you can integrate the LHS by parts. For

2) y'=y²-4y

you also use the fact that the equation is separable. The integration can be done by partial fractions.

 

[SVXX]

In the first question, y/(y-1) = (y - 1 + 1)/(y - 1) = 1 + 1/(y - 1). The integral of this would be y + log(y - 1).
In the second question, you can either complete the square or separate the integrand by partial fractions. By partial fracs, the LHS dy/(y)(y - 4) is separated into => (dy/4y) - (dy/4(y - 4)) which can be easily integrated.

 

[simba_]

For

y/(y-1) dy = 1/t dt

you can integrate the LHS by parts. For

2) y'=y²-4y

you also use the fact that the equation is separable. The integration can be done by partial fractions.

i got the first one... thanks.

for the second one i have to find a way to integrate 1/(y²-4y) dy I am lost as to how i do this, I am assuming its but integration by parts but i cannot get an answer using this method

 

[Mark44]

i got the first one... thanks.

for the second one i have to find a way to integrate 1/(y²-4y) dy I am lost as to how i do this, I am assuming its but integration by parts but i cannot get an answer using this method

Not integration by parts -- use partial fractions. You want to find constants A and B so that 1/(y²-4y) = A/y + B/(y - 4).

 

[simba_]

Not integration by parts -- use partial fractions. You want to find constants A and B so that 1/(y²-4y) = A/y + B/(y - 4).

ahhh thanks... my brain is still moving a bit slow from last night

 

[HallsofIvy]

Also, I would not use "integration by parts for the first one"- use simple algebra:
$$\frac{y}{y- 1}= \frac{y- 1+ 1}{y- 1}= 1+ \frac{1}{y- 1}$$

 

[Metaleer]

One needs to be careful with these separable variable equations.

For instance, let us consider the second equation,

$$\frac{dy}{dt} = y^2 -4 y.$$​

We always have to remember that division by zero is not a mathematically valid operation. So, when we separate variables to write

$$\frac{dy}{y^2 -4 y} = dt,$$​

we need to state that $$y \neq 0$$ and $$y \neq 4$$. Once we integrate on both sides to get $$y(t)$$ in implicit form (and perhaps be able to solve for it in terms of elementary functions), we need to see what happens when $$y = 0$$ and $$y = 4$$. It is easy to see that both of these are solutions to the differential equation, but are not solutions to the initial value problem at hand.

Let's imagine, however, that the initial value problem was $$y(3) = 4$$, for instance. You'd have two solutions for this IVP - the one corresponding to a particular integration constant, and $$y(t) = 4$$. Uniqueness of the solution doesn't hold.