Question

Use separation of variables to solve the initial value problem: dy/dx=(5-x^2)/(3y+5) and y=-1 when x=0
I'm not sure how to use separation of variables

Answer 1

cross multiply

Answer 2

\(\large \frac{dy}{dx} = \frac{5-x^2}{3y+5}\)


\(\large (3y+5 )dy = (5-x^2)dx\)

Answer 3

integrate now

Answer 4

\(\large \frac{dy}{dx} = \frac{5-x^2}{3y+5}\)


\(\large (3y+5 )dy = (5-x^2)dx\)


\(\large \int (3y+5 )dy = \int (5-x^2)dx\)

Answer 5

Thanks! I get the separation of variable now, I have one more question, when I integrate, I end with : (3y^2)/2 +5y=5x-(x^3/5) and I'm not sure how to solve for y

Answer 6

before solving "y", find the integration constant C, by using the initial conditions

Answer 7

when u do integration, the constant always pops up

Answer 8

you should get :

(3y^2)/2 +5y=5x-(x^3/3) + C

Answer 9

right ?

Answer 10

when I solve for (0,-1) I get 3(-1)/2+5(-1)=0+c
c=(-7/2)

*I guess my question with this is if I'm allowed to just input the c on one side? and if my method works

Answer 11

correct !

Answer 12

(3y^2)/2 +5y=5x-(x^3/3) - 7/2

Answer 13

next, solve "y" if u want

Answer 14

or u may leave the solution in implicit form like above

Answer 15

1) Don't ever "cross multiply". Multiply bu things that are non zero and switch to the differential definition of the derivative. This gives: \(y \ne -5/3\) and then the integral expression you have created.

2) You seem to be struggling with your algebra. Try all that, again.

\(\dfrac{3}{2}y^{2} + 5y = 5x - \dfrac{1}{3}x^{3} + C\) -- This is the expected result.

3) Use the initial values given to find C. \(\dfrac{3}{2}(-1)^{2} + 5(-1) = 5(0) - \dfrac{1}{3}(0)^{3} + C\). This is find, since we lose nothing by putting \(C\) on one side or \(C_{1}\) and \(C_{2}\) on either side.

This suggests \(3/2 - 5 = C = -7/2\)

Giving: \(\dfrac{3}{2}y^{2} + 5y = 5x - \dfrac{1}{3}x^{3} - \dfrac{7}{2}\)

4) Who said you had to solve for \(y\)? If you REALLY have to, I'm sure you've solved a Quadratic Equation, before. Give it a go.

Answer 16

depends on wat ur professor wants

Answer 17

awesome! thanks so much! and I think I can leave it in implicit form, at least I hope

Answer 18

np :) if u r forced to solve, just use the quadratic formula