Solution to Initial Value Problem with Separable Equations | Diffy Q Homework

[Jamin2112]

Homework Statement

Find the solution to each of the following initial value problems in explicit form.

(a) y'=(2x)/(y+x²y), y(0)=-2

Homework Equations

Uhhhhh I suppose it would be relevant to mention that an equation in the form dy/dx = g(x)h(y) is separable

The Attempt at a Solution

y'=(2x)/(y+x²y)=[1/y][2x/(1+x²)]

==> y dy=2x/(1+x²) dx
==> y²/2 + C₁ = ln|1+x²| + C₂
==> y²/2 = ln|1+x²| + C (C = C₂ - C₁, just to combine into one constant)

==> (-2)²/2 = ln|1+0²| + C
==> C = 2
==> y² = 2ln|1+0²| + 4
==> y = +/- √(2ln|1+x²| + 4)

That just seems a little too complicated an answer.

 

[Jamin2112]

I just don't understand the whole procedure. If I had put C on the left hand side of the equals sign, I would've came up with a different answer.

 

[rs1n]

I just don't understand the whole procedure. If I had put C on the left hand side of the equals sign, I would've came up with a different answer.

How so? Leaving C on the left simply makes C=-2 as opposed to 2.

 

[Jamin2112]

How so? Leaving C on the left simply makes C=-2 as opposed to 2.

which would change the initial value problem

 

[rs1n]

which would change the initial value problem

No it wouldn't -- not if C is on the left. Are you sure you aren't mixing C's? Had you left C on the left, then you wouldn't be using the equation above (in which C is on the right).

 

[Jamin2112]

No it wouldn't -- not if C is on the left. Are you sure you aren't mixing C's? Had you left C on the left, then you wouldn't be using the equation above (in which C is on the right).

Got'cha. Do you think the rest of the problem is right? I'm getting quite a few answers with "plus or minus" in this homework.

 

[rs1n]

Seems like you understand separable DEQs to me. It also doesn't hurt to check your answers by seeing if the answer satisfies the original DEQ.

 

[Jamin2112]

Seems like you understand separable DEQs to me. It also doesn't hurt to check your answers by seeing if the answer satisfies the original DEQ.

Horrible idea. Square roots and natural logs, combined, are a b***h to integrate.