Please help solve ODE questions for DE newbie

1. y^{2}(dx/dy) + xy -4y^{2 }=1

I try to simplify to

(dx/dy) + x/y -4 = 1/y^{2 }(dy/dx)^{ = }y^{2} - x/y + 1/4

I don't know how to go next (I try to use integrating factor but it's not working)

2. dy/dx = y(y+3)

(dy/dx)/y(y+3) = 1

Integral (dy/dx)/y(y+3) = Integral 1 dx

then I get

1/3 In(y) - 1/3 In(y+3) = x + c

(I'm not sure about this) My calculus is rusty. I don't remember how to get this number I just remember the formula

Someone please explain or correct. :(

Then I'm not sure how to get

y = .......

Re: Please help solve ODE questions for DE newbie

Quote:

Originally Posted by

**angelme** 1. y^{2}(dx/dy) + xy -4y^{2 }=1

I try to simplify to

(dx/dy) + x/y -4 = 1/y^{2 }(dy/dx)^{ = }y^{2} - x/y + 1/4

I don't know how to go next (I try to use integrating factor but it's not working)

2. dy/dx = y(y+3)

(dy/dx)/y(y+3) = 1

Integral (dy/dx)/y(y+3) = Integral 1 dx

then I get

1/3 In(y) - 1/3 In(y+3) = x + c

(I'm not sure about this) My calculus is rusty. I don't remember how to get this number I just remember the formula

Someone please explain or correct. :(

Then I'm not sure how to get

y = .......

For 1) the integrating factor should work.

The integrating factor that I get is $\displaystyle \frac{1}{y}$ If you don't get this integrating factor post your work so we can see what went wrong.

For 2)

Multiply by 3 to get

$\displaystyle \ln(y)+\ln(y+3)=3x+\hat{c}$

$\displaystyle \ln(y(y+3))=3x+\hat{c} \iff y(y+3)=Ae^{3x}$

This is quadratic in y. If you really want to isolate y you can use the quadratic formula.

Re: Please help solve ODE questions for DE newbie

For #1, let me use w instead of x (w=x). It hurts my eyes otherwise.

$\displaystyle y^2w' + yw - 4y^2 = 1$.

$\displaystyle w' + \frac{1}{y}w = 4 + y^{-2}$ (Doing this isn't ideal, since y=0 now becomes a bigger problem. But it's a minor concern.)

Yes, the integrating factor is based on 1/y, so is actually $\displaystyle e^{\int \frac{dy}{y}} = e^{ln(|y|)+c} = A|y|$, so use $\displaystyle y$.

With practice, that's "seeable" in the initial problem, $\displaystyle y^2w' + yw$. Here it amounts to observing that $\displaystyle yw' + w = (yw)'$.

So only divide both sides of $\displaystyle y^2w' + yw = 1 + 4y^2$ by just one power of $\displaystyle y$ to get:

$\displaystyle yw' + w = y^{-1} + 4y$, so $\displaystyle (yw)' = y^{-1} + 4y$. Then you're in business.

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For #2, is it the partial fractions that you're unsure of? It works like this:

Find $\displaystyle \int \frac{dy}{(y)(y+3)}$.

Algebra: $\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3}$ for some constants $\displaystyle A$ and $\displaystyle B$ (it's the opposite of finding common demoninators.)

It's then straightforward to find $\displaystyle A$ and $\displaystyle B$:

$\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3}$ implies (clear demoninators)

$\displaystyle 1 = A(y+3) + By$, implies $\displaystyle 1 = (A+B)y+3A$ for all y, implies $\displaystyle A+B = 0, 3A = 1$.

Thus $\displaystyle A = \frac{1}{3}, B= \frac{-1}{3}$.

Thus $\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3} = \frac{\frac{1}{3}}{y} + \frac{\frac{-1}{3}}{y+3}$

Thus $\displaystyle \int \frac{dy}{(y)(y+3)} = \int \left( \frac{\frac{1}{3}}{y} + \frac{\frac{-1}{3}}{y+3} \right) dy$

$\displaystyle = \frac{1}{3} \int \frac{1}{y} dy + \frac{-1}{3} \int \frac{1}{y+3} dy$

$\displaystyle = \frac{1}{3} ln|y| - \frac{1}{3} ln|y+3| + C$

(And you could simplify more: $\displaystyle = ln\left( \left| \frac{y}{y+3}\right|^{1/3} \right) + C$)