# Thread: Need help with differential equations that have initial conditions?

1. ## Need help with differential equations that have initial conditions?

I'm starting to work with differential equations and I'm having trouble with them when they have initial conditions. For these two: Did I solve them correctly even with the conditions? I don't think I did but I tried :/ What might I have done wrong? (For the second one I chose the 3rd option but it's not showing up)

2. ## Re: Need help with differential equations that have initial conditions?

No, you don't have either of them correct. It's impossible to say why because you didn't show what you did.
Both separate variables as the text says. What did you get when you separated x and y?

What did you get when you integrated?

3. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by HallsofIvy
No, you don't have either of them correct. It's impossible to say why because you didn't show what you did.
Both separate variables as the text says. What did you get when you separated x and y?

What did you get when you integrated?
For the first one after seperating I had ydy=2x^3dx After integrating that I had 1/2y^2=1/2x^4 But then I wasn't sure how to use the initial condition.

The second one I did the same way and got confused with the initial condition. I ended up with 1/2y^2 on one side and 1/2(x+1)^2 on the other, and wasn't sure what to do with y(1)=-2 I feel really stupid that I can't get these.

4. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by canyouhelp
For the first one after seperating I had ydy=2x^3dx After integrating that I had 1/2y^2=1/2x^4 But then I wasn't sure how to use the initial condition.
after integrating you have $\frac{1}{2}y^2=\frac{1}{2}x^4 + C$

C is a constant of integration and it's critical that you remember to include it.

You solve for the value of C using your initial condition.

Plugging the condtion $y(1)=2$ in you get

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

$2=\frac{1}{2}+C$

$C=\frac 3 2$ so

$\frac 1 2 y^2 = \frac 1 2 x^4 + \frac 3 2$

$y^2 = x^4 + 3$

which is seen to be choice 3.

You can work out #9.

5. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by romsek
after integrating you have $\frac{1}{2}y^2=\frac{1}{2}x^4 + C$

C is a constant of integration and it's critical that you remember to include it.

You solve for the value of C using your initial condition.

Plugging the condtion $y(1)=2$ in you get

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

$2=\frac{1}{2}+C$

$C=\frac 3 2$ so

$\frac 1 2 y^2 = \frac 1 2 x^4 + \frac 3 2$

$y^2 = x^4 + 3$

which is seen to be choice 3.

You can work out #9.
For #9 I tried it again and when I plugged in (1,-2) I got option 2. Because I had 2=2+0 So C would be 0. Right?

6. ## Re: Need help with differential equations that have initial conditions?

No that's not correct.

You need to start posting your work.

7. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by romsek
No that's not correct.

You need to start posting your work.
But how is that not right? I plugged 1,-2 into 1/2y^2=1/2(x+1)^2+C Which is 2=2+C, So wouldn't C be 0 and it would be B? If not I integrated it wrong and plugging in the same (1,-2) it would have to be A? But I thought after separating and integrating the equation was 1/2y^2=1/2(x+1)^2?

8. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by canyouhelp
But how is that not right? I plugged 1,-2 into 1/2y^2=1/2(x+1)^2+C Which is 2=2+C, So wouldn't C be 0 and it would be B? If not I integrated it wrong and plugging in the same (1,-2) it would have to be A? But I thought after separating and integrating the equation was 1/2y^2=1/2(x+1)^2?
You did not separate the variables correctly. You had:

$\dfrac{dy}{dx} = \dfrac{1+x}{xy}$

Multiplying both sides by $ydx$ gives:

$ydy = \dfrac{1+x}{x}dx$

You integrated the left hand side correctly, but not the night. Note:

$\dfrac{1+x}{x} = \dfrac{1}{x} + \dfrac{x}{x} = x^{-1} + 1$

Integrating that does not give $\dfrac{1}{2}(x+1)^2+C$.

Note how easily I was able to help when you actually posted your work.

9. ## Re: Need help with differential equations that have initial conditions?

Originally Posted by SlipEternal
You did not separate the variables correctly. You had:

$\dfrac{dy}{dx} = \dfrac{1+x}{xy}$

Multiplying both sides by $ydx$ gives:

$ydy = \dfrac{1+x}{x}dx$

You integrated the left hand side correctly, but not the night. Note:

$\dfrac{1+x}{x} = \dfrac{1}{x} + \dfrac{x}{x} = x^{-1} + 1$

Integrating that does not give $\dfrac{1}{2}(x+1)^2+C$.

Note how easily I was able to help when you actually posted your work.
Oh! Okay, thank you so much. When I integrate x^-1+1 I get ln(x)+x+C. So 1/2y^2=ln(x)+x+C with (1,-2) plugged in is 2=0+1+C. So C must = 1, making it option A?

yes