Solving First Order Differential Equation with Initial Value x(0)=2

[rugapark]

I'm trying to solve this firrst order diff. equation, where I'm given the initial value, x(0)=2$$\frac{dx}{dt}=\frac{3x+4}{\sqrt{t}}$$

$$\frac{dx}{3x+4}=\frac{1}{\sqrt{t}}dt$$

$$\int\frac{1}{3x+4}dx=\int\frac{1}{\sqrt{t}}dt$$

$$ln(3x+4)=ln(t^{\frac{1}{2}})$$

this is as far as I got, do I sub in x(0)=2 into the LHS? if not, could i have some pointers to help carry on?

cheers.

 

[Hootenanny]

I'm trying to solve this firrst order diff. equation, where I'm given the initial value, x(0)=2


$$\frac{dx}{dt}=\frac{3x+4}{\sqrt{t}}$$

$$\frac{dx}{3x+4}=\frac{1}{\sqrt{t}}$$

$$\int\frac{dx}{3x+4}=\int\frac{1}{\sqrt{t}}$$

$$ln(3x+4)=ln(t^{\frac{1}{2}}$$

Your solution looks good up until the final line. You may want to re-check your integrals.

 

[rugapark]

[edit] - there should be 'dt' s on the RHS in the 2nd and 3rd line of work

 

[HallsofIvy]

$$\int \frac{dt}{\sqrt{t}}= \int t^{-1/2}dt$$
is NOT $ln(t^{1/2})$!

 

[salman213]

also don't forget your constant that comes from the integration

 

[rugapark]

oh right, my mistake lol

so, the last line should be

$$ln(3x+4)=\frac{1}{2}ln(t^{\frac{1}{2}})$$

I have no idea where to go from here with the initial value that I was given x(0)=2! would I sub in x=2 into the equation?

 

[Dick]

oh right, my mistake lol

so, the last line should be

$$ln(3x+4)=\frac{1}{2}ln(t^{\frac{1}{2}})$$

I have no idea where to go from here with the initial value that I was given x(0)=2! would I sub in x=2 into the equation?

As other people have been trying to tell you, the integral of t^(-1/2) DOES NOT involve a log. It's just power law. And I still don't see a constant of integration.