how to expres in terms of y

**lord12** · September 21st 2009, 12:56 AM

dy/dt = t/2y+1. Find the general solution.

I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?

**Prove It** · September 21st 2009, 01:15 AM

Originally Posted by lord12

dy/dt = t/2y+1. Find the general solution.

I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?

Complete the square.

$y^2 + y + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{2}t^2 + C$

$\left(y + \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{1}{2}t^2 + C$

$\left(y + \frac{1}{2}\right)^2 = \frac{1}{2}t^2 + \frac{1}{4} + C$

$y + \frac{1}{2} = \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}$

$y= \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}$ .

**mr fantastic** · September 21st 2009, 02:56 AM

Originally Posted by Prove It

[snip]

$y + \frac{1}{2} = {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}$

$y= {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}$ .

..

**Prove It** · September 21st 2009, 03:25 AM

Originally Posted by mr fantastic

..

Yes - I was hoping that would be implied...

**lord12** · September 21st 2009, 03:39 AM

dy/dt = t/2y+1

y = sqrt(1/2t^2 + 1/4 +k)

y(1) = 3/2

**Prove It** · September 21st 2009, 04:19 AM

Originally Posted by lord12

dy/dt = t/2y+1

y = sqrt(1/2t^2 + 1/4 +k)

y(1) = 3/2

First, your equation for $y$ is incorrect.

$y = \pm\sqrt{\frac{1}{2}t^2 + \frac{1}{4} + C} - \frac{1}{2}$ .

If $y(1) = \frac{3}{2}$ , then

$\frac{3}{2} = \pm \sqrt{\frac{1}{2}(1)^2 + \frac{1}{4} +C} - \frac{1}{2}$

$2 = \pm \sqrt{\frac{1}{2} + \frac{1}{4} + C}$

$4 = \frac{3}{4} + C$

$\frac{13}{4} = C$ .

I think you'll find the solution is NOT unique...

$y = \pm \sqrt{\frac{1}{2}t^2+ \frac{1}{4} + \frac{13}{4}} - \frac{1}{2}$

$= \pm \sqrt{\frac{1}{2}t^2 + \frac{7}{2}} - \frac{1}{2}$ .

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