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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

$t(t-4)y'+y=0 \quad y(2)=2\quad 0<t<4$

ok my first step was isolate y'

s

$y'=-\dfrac{y}{t(t-4)}$

not sure what direction to go since we are concerned about an interval

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,886

$t(t-4)y'+y=0 \quad y(2)=2\quad 0<t<4$

ok my first step was isolate y'

s

$y'=-\dfrac{y}{t(t-4)}$

not sure what direction to go since we are concerned about an interval

- Jan 30, 2018

- 729

Well, the problem specifically says "0< t< 4" and y' does not exist at t= 0 and t= 4.

- Thread starter
- #3

- Jan 31, 2012

- 2,886

how does y(2)=2 fit into this

doesn't that give us specific y interval

doesn't that give us specific y interval

- Aug 30, 2012

- 1,178

You could always do "brute force" if you can't figure out a work around. Boundary conditions get rid of integration constants. Solve the differential equation. What is y(t)? What does that tell you about the solution interval(s)?how does y(2)=2 fit into this

doesn't that give us specific y interval

-Dan

- Jan 30, 2018

- 729