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Differential equation advice please

is it possible to solve

dydt=y−x\frac{dy}{dt} = \sqrt y - \sqrt x ?

the bear

:badger:
Reply 1
Avatar for the bear
the bear
OP
20
i have plotted a gradient field using x and y axes...

how should the solution curves from the field be used if the formula is dy/dt = x^0.5 - y^0.5 rather than dy/dx = x^0.5 - y^0.5 ?

the baer

:confused::confused:
Reply 2
Avatar for McMannBean
McMannBean
How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

I've not actually tried it out on paper mind, so it may not work. But hope it helps
Original post by McMannBean
How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

I've not actually tried it out on paper mind, so it may not work. But hope it helps

You can't use the integrating factor method because the ODE is non-linear.
Reply 4
Avatar for McMannBean
McMannBean
let's try a substitution then:

Let U = sqrt(y)

so we end up with 2U*dU/dt - U = const

=> dU/dt - const*U^-1 = 1

This should be solveable
Reply 5
Avatar for nuodai
nuodai
17
Original post by McMannBean
let's try a substitution then:

Let U = sqrt(y)

so we end up with 2U*dU/dt - U = const

=> dU/dt - const*U^-1 = 1

This should be solveable


How do we know that xx is independent of tt? [If it's not then we certainly can't assume it's constant.]

I think we lack enough information to be able to find a solution here. Is there more to the question?
Reply 6
Avatar for the bear
the bear
OP
20
Thanks so much guys for those helpful suggestions. As Nuodai said it would be better if x were a function of t .

it may be that the function is quadratic:

x = ( k - t )^2

where k is a constant

then the equation would become:

dy/dt = y^0.5 - ( k - t )

& then the tangent field with y and t axes would be appropriate: here the horizontal axis is t: k is 20
5621272199_12c08b9481_z.jpg352.6KB

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