Question

Solve the first order ODE: y'= ((y^2)x +x)/(y(x^2)+y)

Answer 1

Is it \[ y \prime = \frac{y^2x + x}{y(x^{2}) + y} \]Maybe factoring out x from the numerator and y from the denominator to get: \[ y \prime = \frac{x(y^2 + 1)}{y(x^{2} + 1)} \]. Divide both sides by (y^2 + 1) and multiply both sides by y to get:\[\frac{yy \prime}{y^2 +1} = \frac{x}{x^{2} + 1} \]. The integral will be \[ \int \frac{y}{y^2 + 1}dy = \int \frac{x}{x^2 + 1}dx \]

Answer 2

Yup, thats everything i have so far.

Answer 3

But i have to substitute now or something. thats where im stuck

Answer 4

Both integrals can be solved by using u = denominator. For the y one, we get: u = y^2 + 1, du = 2ydy, so we have to multiply by 2/2. The integral becomes:\[\frac{1}{2} \int\limits \frac{du}{u}\]

Answer 5

1/2 ln(y^2+1) +c = 1/2 ln(x^2 +1)+c

Answer 6

Yeah. Do you neead an explicit solution?

Answer 7

need*

Answer 8

whats that?

Answer 9

What you got is a solution for the DE already, but it's in function of ln of y. Many times, you have to write it as y = something. Whenever you write it as y = something that's an explicit function of y.

Answer 10

Ohhhh. Yea, i need it in y=

Answer 11

So, we have:\[\frac{1}{2}\ln{(y^2 + 1)} + c_1 = \frac{1}{2} \ln{(x^2 + 1)} + c_2\]Subtract c1 from both sides and rewrite it as\[\frac{1}{2}\ln{(y^2 + 1)} = \frac{1}{2}\ln(x^2 +1) + c\]Multiply both sides by 2, we get:\[\ln{(y^2+1)} = \ln{(x^2 + 1)} + C\]

Answer 12

Take the e base exponent on both sides:\[\LARGE e^{\ln(y^2 + 1)} = e^{\ln(x^2 + 1) + C}\]Which is equal to\[y^2 + 1 = e^C (x^2+1)\]Subtract one from both sides and square root to get:\[y = \pm \sqrt{K(x^2 + 1) - 1}\]with K = e^C. In order to pick a solution, we need some initial value.

Answer 13

Alritey. Well since there is no initial value, do I leave it at y=.... or do I have to put y=.... AND K=e^c??

Answer 14

This can be also written as \[ y = \pm \sqrt{Kx^2 - C} \] with C = K -1.

Answer 15

K = e^c is not a solution. Both positive and negative values of the square root are solutions. I think it's better for you to leave as y = ...

Answer 16

i.e., y = sqrt(K(x^2 + 1) - 1) and y = - sqrt(K(x^2 + 1) - 1) are solutions

Answer 17

Great!! Thaaanks!! Do you know how to do arc lengths of polar curves?