Question

y=(1+p)x+p^2
where p =dy/dx
please solve it .
kuch b lago bas mere dost answer nikal do homogenous jo kuch b lago.

ANS. x = 2(1-p)+ce^-p
y = 2-p^2 + (1+p)ce^-p


1 REQUEST
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Answer 1

Answer:

Example: Solve dydx = x2 + y2xy

Can we get it in F( yx ) style?

Start with:x2 + y2xy

Separate terms:x2xy + y2xy

Simplify:xy + yx

Reciprocal of first term:( yx )−1 + yx

Yes, we have a function of (y/x).

So let's go:

Start with:dydx = ( yx )−1 + yx

y = vx and dydx = v + xdvdx:v + xdvdx = v−1 + v

Subtract v from both sides:xdvdx = v−1

Now use Separation of Variables:

Separate the variables:v dv = 1x dx

Put the integral sign in front:∫v dv = ∫1x dx

Integrate:v22 = ln(x) + C

Then we make C = ln(k):v22 = ln(x) + ln(k)

Combine ln:v22 = ln(kx)

Simplify:v = ±√(2 ln(kx))

Now substitute back v = yx

Substitute v = yx:yx = ±√(2 ln(kx))

Simplify:y = ±x √(2 ln(kx))

And we have the solution.