Answer to Question #241053 in Differential Equations for Randal Rodriguez

Question

solve the given equations using exact differential

equations.

6. (cos x ∙ cos y − cot x)dx − (sin x ∙ sin y)dy = 0

7. x(3xy − 4y3 + 6)dx + (x3 − 6x2y2 − 1)dy = 0, when x = 2, y = 0

8. 2xydx + (y2 + x2)dy = 0

9. (xy2 + y − x)dx + x(xy + 1)dy = 0, when x = 1, y = 1

|10. (1 − xy)−2dx + [y2 + x2(1 − xy)−2]dy = 0, when x = 3, y = 4

Answer 1

$\displaystyle (\cos x. \cos y- \cot x)dx - (\sin x \sin y)dy= 0\\ \text{Let $M = \cos x. \cos y- \cot x$ and $N = -\sin x \sin y$}\\ \text{Let $F(x,y) = \int (-\sin x \sin y)dy$}\\ \implies F(x,y) = \sin x \cos y + h(y)\\ \text{Differentiating and equating it to M, we have}\\ \cos x \cos y +h'(x) = \cos x \cos y -\cot x\\ \implies h'(x) = -\cot x\\ h(x) = -\ln \sin x\\ \therefore F(x,y) = \sin x \cos y -\ln \sin x\\ \text{The solution to the exact differential equation is}\\ \sin x \cos y -\ln \sin x=c\\ 2. (3x^2y-4y^3+6)dx - (x^3-6x^2y^2-1)dy= 0\\ \text{Let $M = 3x^2y-4y^3+6$ and $N = x^3-6x^2y^2-1$}\\ \text{Let $F(x,y) = \int (x^3-6x^2y^2-1)dy$}\\ \implies F(x,y) = x^3y-2x^2y^3-y + h(y)\\ \text{Differentiating and equating it to M, we have}\\ 3x^2y-4y^3+h'(x) = 3x^2y-4y^3+6\\ \implies h'(x) = 6\\ h(x) = 6x\\ \therefore F(x,y) = x^3y-2x^2y^3-y +6x\\ \text{The solution to the exact differential equation is}\\ x^3y-2x^2y^3-y +6x=c\\ \text{When x = 2, y=0, hence c = 12}\\ \implies x^3y-2x^2y^3-y +6x=12 3. (2xy)dx - (y^2+x^2)dy= 0\\ \text{Let $M = 2xy$ and $N = y^2+x^2$}\\ \text{Let $F(x,y) = \int (y^2+x^2)dy$}\\ \implies F(x,y) = \frac{y^3}{3}+x^2y + h(x)\\ \text{Differentiating and equating it to M, we have}\\ 2xy +h'(x) = 2xy\\ \implies h'(x) = 0\\ h(x) = c\\ \therefore F(x,y) = \frac{y^3}{3}+x^2y+c\\ \text{The solution to the exact differential equation is}\\ \frac{y^3}{3}+x^2y+c=c\\ \implies \frac{y^3}{3}+x^2y+c=0 4. (xy^2+y-x)dx - (x^2y+x)dy= 0\\ \text{Let $M = xy^2+y-x$ and $N = x^2y+x$}\\ \text{Let $F(x,y) = \int (x^2y+x)dy$}\\ \implies F(x,y) = \frac{x^2y^2}{2}+xy + h(x)\\ \text{Differentiating and equating it to M, we have}\\ xy^2+y+h'(x) = xy^2+y-x\\ \implies h'(x) = -x\\ h(x) = -\frac{x^2}{2}\\ \therefore F(x,y) = x^3y-2x^2y^3-y +6x\\ \text{The solution to the exact differential equation is}\\ \frac{x^2y^2}{2}+xy-\frac{x^2}{2}=c\\ \text{When x = 1, y=1, hence c = 1}\\ \implies \frac{x^2y^2}{2}+xy-\frac{x^2}{2}=1\\ 5. (1-xy-2)dx + (y^2+x^2-x^3y-2)dy\\ \text{The equation isn't exact as}\\ M_y =-x, N_x=2x-3x^2y\\ M_y \neq N_x$