# Differential Equation Questions

#### 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!

#### AllanCuz

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

#### Prove It

MHF Helper
Who says that's wrong? Looks A-okay to me
If $$\displaystyle y^2 = x^2 + 25$$ then surely $$\displaystyle y = \pm\sqrt{x^2 + 25}$$.

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#### AllanCuz

If $$\displaystyle y^2 = x^2 - 25$$ then surely $$\displaystyle y = \pm\sqrt{x^2 - 25}$$.
huh?

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

$$\displaystyle 25 = C$$

$$\displaystyle y^2 = x^2 + 25$$

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

?

#### Prove It

MHF Helper
huh?

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

$$\displaystyle 25 = C$$

$$\displaystyle y^2 = x^2 + 25$$

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

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

#### break

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

#### mr fantastic

MHF Hall of Fame
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 $$\displaystyle y = {\color{red}-} \sqrt{x^2 + 25}$$, as is easily confirmed by substitution. Note that this solution satisfies y(0) = -5 ....

AllanCuz