# Math Help - Differential Equation Questions

1. ## Differential Equation Questions

I'm stuck on this problem

Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

first i move y and x around and get y(dy)=x(dx)

then get the integral of both

y^2/2 = x^2/2 + C

multiply both sides by 2,

y^2 = x^2 + C

find C by plugging in 0 for X and -5 for Y

-5^2 = 0 + C

25 = C

I plug C into the regular equation and

then my final answer is y=sqrt( x^2 +25 )

but it is wrong. What am i doing wrong? thanks!

2. Originally Posted by break
I'm stuck on this problem

Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

first i move y and x around and get y(dy)=x(dx)

then get the integral of both

y^2/2 = x^2/2 + C

multiply both sides by 2,

y^2 = x^2 + C

find C by plugging in 0 for X and -5 for Y

-5^2 = 0 + C

25 = C

I plug C into the regular equation and

then my final answer is y=sqrt( x^2 +25 )

but it is wrong. What am i doing wrong? thanks!
Who says that's wrong? Looks A-okay to me

3. Originally Posted by AllanCuz
Who says that's wrong? Looks A-okay to me
If $y^2 = x^2 + 25$ then surely $y = \pm\sqrt{x^2 + 25}$.

4. Originally Posted by Prove It
If $y^2 = x^2 - 25$ then surely $y = \pm\sqrt{x^2 - 25}$.
huh?

$y^2 = x^2 + C$ with the IC $y(0) = - 5$

$25 = C$

$y^2 = x^2 + 25$

$y = \sqrt{ x^2 + 25}$

?

5. Originally Posted by AllanCuz
huh?

$y^2 = x^2 + C$ with the IC $y(0) = - 5$

$25 = C$

$y^2 = x^2 + 25$

$y = \sqrt{ x^2 + 25}$

?
Sorry, I put a minus where there should have been a plus. The point is, your work was fine, you just had to make $y$ the subject.

6. This is one of my online homework problems, apparently y=-sqrt(x^2+25) is correct. but i still don't understand why @_@.

7. Originally Posted by break
I'm stuck on this problem

Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

first i move y and x around and get y(dy)=x(dx)

then get the integral of both

y^2/2 = x^2/2 + C

multiply both sides by 2,

y^2 = x^2 + C

find C by plugging in 0 for X and -5 for Y

-5^2 = 0 + C

25 = C

I plug C into the regular equation and

then my final answer is y=sqrt( x^2 +25 )

but it is wrong. What am i doing wrong? thanks!
The correct answer is $y = {\color{red}-} \sqrt{x^2 + 25}$, as is easily confirmed by substitution. Note that this solution satisfies y(0) = -5 ....