Determine wheather the given equation is exact. If it is exact,Solve it?

Question

(a) (2 ln y +y^ 2/ x ^2) dx +( x^2 / y - 2 y ln x )

(b) 2ty / t^2 +1 - 2t - ( 2 - ln ( t^2 + 1 )) dy/dt =0

(c) Find the solution of the linear differential equation
x dy/dx +2=x^2 -x +1

Answer 1

The form M(x,y) dx + N(x,y) dy = 0 is exact iff ∂M/∂y = ∂N/∂x

If this condition is met, then M = ∂P/∂x and N = ∂P/∂y for some P(x,y), and the equation has the form

∂P/∂x dx + ∂P/∂y dy = 0

which may be written as dP(x,y)=0 and whose solution is P(x,y)=c where

P = ∫ M(x,y) dx = ∫ N(x,y) dy (a sort of "partial integration").

For (a) (which I assume is (2 ln y +y²/x²) dx +(x²/y - 2y ln x) dy = 0)

∂M/∂y = 2/y - 2y/x² and ∂N/∂x = 2x/y - 2y/x

so the equation is not exact.

For (b), which may be written

(2ty/(t²+1) - 2t) dt + (ln(t²+1) - 2) dy = 0, we have

∂(2ty/(t²+1) - 2t)/∂y = 2t/(t²+1) and ∂(ln(t²+1) - 2)/∂t = 2t/(t²+1)

so this equation is exact.

∫ M(t,y) dt = ∫ (2ty/(t²+1) - 2t) dt = y ln(t²+1) - t² + g(y) for arbitrary function g(y) (hence the notion of "partial integration"). Since this is to be P(t,y), and we want

∂P/∂y = N(t,y) = ln(t²+1) - 2, we have to have

∂(y ln(t²+1) - t² + g(y))/∂y = ln(t²+1) - 2, so

ln(t²+1) + g'(y) = ln(t²+1) - 2, so

g'(y) = -2, which implies that g(y) = -2y

Therefore, P(t,y) = y ln(t²+1) - t² - 2y and the solution is

y ln(t²+1) - t² - 2y = c

For practice, you should find P(t,y) by integrating (ln(t²+1) - 2) with respect to y. You should, of course, get the same P(t,y).

I think you have a typo in (c); if not, subtract 2 from both sides, divide both sides by x, and integrate both sides with respect to x.

Answer 2

Part (c) is the only one that is even answerable; the other two are ambiguously written, and the first one doesn't even make sense.

The third one is not exact, for, taking M(x,y) = x^2 - x - 1 and N(x,y) = -x, it is clear that dM/dy does not equal dN/dx. It is easily solvable, however. Just move the 2 on the left over to the right, divide by x, and integrate. You will have a family of solutions for all x not equal to 0.

dy/dx = x - 1 - 1/x
y = 1/2 x^2 - x - log|x| + c

Since y becomes very large as x approaches zero, there is no solution on any interval including x = 0.