1. ## Differential Equations help please!

hi, im having problems with these...

find the general solutions for:

t^2 dx/dt = x +1

and

y^1/2 dy/dx = y/x

if someone could go step by step on those i would be very happy!

many thanks

2. Originally Posted by united_legend
hi, im having problems with these...

find the general solutions for:

t^2 dx/dt = x +1

and

y^1/2 dy/dx = y/x

if someone could go step by step on those i would be very happy!

many thanks
Theser are SDEs(seperable differential equations)...for example the first one $\displaystyle t^2\frac{dx}{dt}=x+1$...sepearing we get $\displaystyle \frac{1}{x+1}dx=\frac{dt}{t^2}$...integrating both sides we get $\displaystyle ln|x+1|=\frac{-1}{t}+C$ now solving for the function x which is what we want we get $\displaystyle x=Ce^{\frac{-1}{t}}-1$ where C is a constant..

3. Originally Posted by united_legend
hi, im having problems with these...

find the general solutions for:

t^2 dx/dt = x +1

and

y^1/2 dy/dx = y/x

if someone could go step by step on those i would be very happy!

many thanks
Now you should submit your work to see if you are doing it correctly but the answer to 2 is $\displaystyle y=\bigg(\frac{\ln(x)}{2}+C\bigg)^2$

4. Hellooo,

Originally Posted by Mathstud28
integrating both sides we get [tex]ln|x+1|=\frac{t^3}{3}+C[/math
Ow, really ?

I thought that 1/tē was the derivative for -1/t

5. i still cant do it... :/

sorry if i sound completely stupid, but how do you seperate it to get that?

cheers!

6. Originally Posted by Moo
Hellooo,

Ow, really ?

I thought that 1/tē was the derivative for -1/t
Ehhhhhhhhhhh.......I have gotten to the point where I dont care if you mock me because I think you know it isnt because I dont understand

7. Originally Posted by united_legend
i still cant do it... :/

sorry if i sound completely stupid, but how do you seperate it to get that?

cheers!
You want to get the dx's with the x's and the dt's with the dt.....and then integrate as you normally would then solve for teh function they ask you to(the function you are supposed to solve for is the whatever letter is on the top of the fraction...example $\displaystyle \frac{dy}{dx}$ it would be solve for y)

8. t^2 dx/dt = x+1

integrate 1/x+1 dx = integrate t^-2 dt

?

which is ln x+1 = -t^-1

:S wat am i doing wrong?

9. No, this is correct

Don't forget the constant of integration !

(moo)

10. Originally Posted by united_legend
t^2 dx/dt = x+1

integrate 1/x+1 dx = integrate t^-2 dt

?

which is ln x+1 = -t^-1

:S wat am i doing wrong?
Yeah if you will notice I accidentally integrated $\displaystyle t^2$ instead of $\displaystyle \frac{1}{t^2}$...but as you will also notice...I corrected myeslef awhile ago...you are doing it correctly

11. Oh right, thanks

and for the 2nd one i get..

y^-2 dy/dx = y/x

integrate y^1/2 /y dy = integrate 1/x dx

integrate y^-1/2 dy = integrate x^-1 dx

= 1/2 y^1/2 = x

?

i get confused with the last bit, you add 1 to the power and divide by the new power dont you?

thanks a lot so far!

12. Originally Posted by united_legend
Oh right, thanks

and for the 2nd one i get..

y^-2 dy/dx = y/x

integrate y^1/2 /y dy = integrate 1/x dx

integrate y^-1/2 dy = integrate x^-1 dx

= 1/2 y^1/2 = x

?

i get confused with the last bit, you add 1 to the power and divide by the new power dont you?

thanks a lot so far!
you have $\displaystyle \sqrt{y}\frac{dy}{dx}=\frac{y}{x}$ dividing both sides by y we get $\displaystyle y^{\frac{-1}{2}}dy=dx\frac{1}{x}$..integrating we get $\displaystyle 2\sqrt{y}=ln|x|+c$ make sense from there?

13. Originally Posted by Mathstud28
you have $\displaystyle \sqrt{y}\frac{dy}{dx}=\frac{y}{x}$ dividing both sides by y we get $\displaystyle y^{\frac{-1}{2}}dy=dx\frac{1}{x}$..integrating we get $\displaystyle 2\sqrt{y}=ln|x|+c$ make sense from there?
i dont no how you get the $\displaystyle 2\sqrt{y}$ bit

cheers

14. Originally Posted by united_legend
i dont no how you get the $\displaystyle 2\sqrt{y}$ bit

cheers
$\displaystyle \int{y^{\frac{-1}{2}}dy}=\frac{y^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}=2y^{\frac{1}{2}}=2\sqrt{y}$