Dy/dx=f(y)+g(x)

By Brad Rodgers (P1930) on Tuesday, November 7, 2000 - 02:55 am :

I am having a bit of trouble finding a method of solving the differential equation of

dy/dx=f(y)+g(x)

How do we solve these sorts of equations?

Thanks,

Brad

By Brad Rodgers (P1930) on Tuesday, November 7, 2000 - 07:49 pm :

Thanks,

I had originally wished to solve the equation of

dy/dx=y+x

I just can't seem to do this methodically (I think that a solution is y=-x-1, but there most probably are other solutions). Any help would be greatly appreciated.

Brad

By Michael Doré (Md285) on Tuesday, November 7, 2000 - 08:01 pm :

Ah well that's much easier than the general case. The nice thing with your example is that the differential equation is linear in y.

You have already found one solution: y = -x - 1 (to see this works just plug in y = -x - 1 to both sides and make sure they balance).

So if you let y₁ = -x - 1 then:

dy₁ /dx = y₁ + x [equation 1]

Now suppose there exists another solution, y such that

dy/dx = y + x [equation 2]

Subtract [1] from [2]:

dy/dx - dy₁ /dx = y - y₁ + x - x

So:

d(y - y₁ )/dx = (y - y₁ )

So the function u = y - y₁ satisfies:

du/dx = u

Therefore u = Ae^x , y - y₁ = Ae^x

y = - x - 1 + Ae^x

And remember that this holds for any y that satisfies the differential equation, so the general solution is:

y = -x - 1 + Ae^x

where A is an arbitrary constant.

Hope this helps,

Michael

By Tom Hardcastle (P2477) on Tuesday, November 7, 2000 - 10:05 pm :

There is also an alternative method which can be used to solve the more general dy/dx = y.f(x) + g(x) which you might be interested in.

If dy/dx + Py = Q where P and Q are functions of x
dy/dx + Py = Q

Now multiply through by

R = e^{\int P dx}

which gives you
R dy/dx + RPy = RQ
Now d(Ry)/dx = R dy/dx + RPy.
So d(Ry)/dx = RQ
So

Ry = \int (R Q) dx

.
So in this case, Q = x and P = -1, so R = e^-x

y e^{- x} = \int (x e^{- x}) dx = - x e^{- x} - e^{- x} + A

y = -x - 1 + Ae^x

Clearly, for this case, the method outlined by Michael is much simpler. But when P is not a constant and the differential equation is linear, this method actually becomes useful.

Tom.

By Kerwin Hui (Kwkh2) on Wednesday, November 8, 2000 - 02:11 pm :

So the general solution to the differential equation

dy / dx + P (x) y = Q (x)

$y = e^{- \int P (x) dx} \int (Q (x) e^{\int P (x') dx'} dx$

Kerwin