Help with a simple differential equation from calc II?

Question

Ok, I'm lost on this section. This is an unassigned HW question I think would help me to figure out if I could see how to do this the question is to find a general form for x'-y^2=2x or dx/dy - y² = 2x and then to show it is really the answer.
I can't figure this out, any help?

Answer 1

Er, I don't know what the above answerer is thinking, but thats complete nonsense. You can't do anything like that with differential equations. (Try differentiating the last thing with respect to x, and see that you don't get anything similar to the original equation).

Anyway, we have x' - 2x = y^2.
This is a first order differential equation, and there are several approaches, depending on how much you know about differential equations.

The first step is to solve x' - 2x = 0.
Thats easy; dx/dy = 2x, so dx/x = 2 dy, so ln|x| = 2y + c, or x = Ce^(2y) for some constant c.

Now, we ignore that part, and concentrate on the original equation.
The easiest next step would be to guess that x = ay^2 + by + c for some a,b,c, and solve.
Then x' = 2ay + b, so substituting in gives:
x' - 2x = y^2
2ay + b - 2ay^2 - 2by - 2c = y^2.
So y^2(-2a) + y(2a-2b) + (b-2c) = y^2.
Equating coefficients, we must have -2a = 1, 2a-2b = 0, b-2c=0.
That tells us a = -1/2, b = -1/2, c = -1/4.
So x = -0.5y^2 - 0.5y - 0.25 is a specific solution.

Now since putting x = Ce^(2y) gives 0 on the left hand side, adding this to our solution won't change the RHS. So the general solution is:

x = -0.5y^2 - 0.5y - 0.25 + Ce^(2y) for any constant C.

Finally, check that works:
x' = -y - 0.5 + 2Ce^(2y).
So x' - y^2 = -y^2 - y - 0.5 + 2Ce^2y = 2x.

Answer 2

no longer which you have been inevitably asserting this, yet just to be sparkling, differential equations isn't organic math, it relatively is rooted in utility and you in no way even coach maximum of what you study. somewhat lots each engineering self-discipline is going onto a linear algebra direction. some end there in actuality, and others pass forward to handle despite they choose of their very own field. As you may locate, your question replaced right into some enormous field, so there is not any set 'what comes next' like there is in severe college. loads of disciplines in some unspecified time interior the destiny take an 'stepped forward math for engineers' direction (physics pupils take an 'stepped forward math for physicists' direction, they are not the comparable) that addressed partial differential equations, remodel tactics, and complicated diagnosis. a lot of human beings additionally decide for to take a risk direction. you're somewhat lots achieved although math-clever once you get via differential equations, the the remainder of the ideas you study as you bypass on your engineering classes.

Answer 3

dx/dy - y^2 = 2x
multiply out by dy so that
dx - y^2dy = 2xdy
Integrate out both sides
I[dx] - I[y^2dy] = 2xI[dy] + constant
x - (1/3)y^3 = 2xy + constant
3x - y^3 = 6xy + constant
Is the what you are looking for?

Answer 4

dx/dy - y² = 2x
dx/dy = y^2 + 2x
dx = y^2dy +2x dy
dx = 2y(dy) + 2x dy
dx/dx =2y dy/dx + 2x dy/dx
1=2y dy/dx + 2x dy/dx
1= dy/dx( 2y+2x)
dy/dx= 1/ (2(x+y))
You should be able to solve for y by integrating 1/(2(x+y))