Q2 TUT 0401 and TUT 0601

[Victor Ivrii]

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
x^2y^3 + x(1 + y^2)y' = 0,\qquad  \mu(x, y) = 1/xy^3.
$$

[Pengyun Li]

(x²y³)+x(1+y²)y'= 0

Let M = x²y³, N = x(1+y²)

M_y = $\frac{d(M)}{dy}$ = 3x²y²

N_x = $\frac{d(N)}{dx}$ = 1+y²

Since M_y ≠ N_x, hence not exact, and thus we need to use the integrating factor μ = 1/(xy³)

Multiply μ on both sides of the original equation,

1/(xy³)x²y³ + 1/(xy³)x(1+y²)y' = 0

Now let M' = 1/(xy³)x²y³ = x,

N' = 1/(xy³)x(1+y²) = y^-3 + y^-1,

M'_y = $\frac{d(M')}{dy}$ = 0,

N'_x = $\frac{d(N')}{dx}$ = 0,

M'_y = N'_x, hence exact now.

There exist φ(x,y) s.t. φ_x = M', φ_y = N'

φ(x,y) = ∫M'dx =$\frac{1}{2}$x² + h(y)

φ_y = h'(y) = N' = y^-3 + y^-1

Thus h'(y) = y^-3 + y^-1, h(y) = -$\frac{1}{2}$y^-2 + ln|y| + C

Therefore, φ(x,y) =$\frac{1}{2}$x² -$\frac{1}{2}$y^-2 + ln|y|= C

[Victor Ivrii]

You need to learn $\LaTeX$ to avoid a shappy html math

And no double-dipping next time (just 1 quiz)