Answer to Question #241052 in Differential Equations for Randal Rodriguez

Question

solve the given equations using exact differential

equations.

1. (6x + y2)dx + y(2x − 3y)dy = 0

2. (y2 − 2xy + 6x)dx − (x2 − 2xy + 2)dy = 0, when x = 1, y = 2

3. v(2uv2 − 3)du + (3u2v2 − 3u + 4v)dv = 0, when u = 1, v = 1

4. (1 + y2 + xy2)dx + (x2y + y + 2xy)dy = 0

5. (w3 + wz2 − z)dw + (z3 + w2z − w)dz = 0, when w = 4, z = 2

Answer 1

$\displaystyle (6x+y^2 )dx - (2xy - 3y^2)dy= 0\\ \text{Let $M = 6x+y^2 $ and $N = 2xy - 3y^2$}\\ \text{Let $F(x,y) = \int (2xy - 3y^2)dy$}\\ \implies F(x,y) = xy^2-y^3 + h(x)\\ \text{Differentiating and equating it to M, we have}\\ y^2 +h'(x) = y^2+6x\\ \implies h'(x) = 6x\\ h(x) = 3x^2\\ \therefore F(x,y) = xy^2-y^3 + 3x^2\\ \text{The solution to the exact differential equation is}\\ xy^2-y^3 + 3x^2=c\\ 2. (y^2-2xy+6x)dx - (x^2-2xy+2)dy= 0\\ \text{Let $M = y^2-2xy+6x $ and $N = -(x^2-2xy+2)$}\\ \text{Let $F(x,y) = \int -(x^2-2xy+2)dy$}\\ \implies F(x,y) = -x^2y+xy^2-2y + h(x)\\ \text{Differentiating and equating it to M, we have}\\ y^2 -2xy+h'(x) = y^2-2xy+6x\\ \implies h'(x) = 6x\\ h(x) = 3x^2\\ \therefore F(x,y) =-x^2y+xy^2-2y+ 3x^2\\ \text{The solution to the exact differential equation is}\\-x^2y+xy^2-2y+ 3x^2=c\\ \text{When x= 1, y=2, c = 1, therefore the particular solution is}\\ -x^2y+xy^2-2y+ 3x^2=-1\\ 3. v(2uv^2-3)du + (3u^2v^3-3u+4v)dv= 0\\ \text{Let $M = v(2uv^2-3) $ and $N = 3u^2v^2-3u+4v$}\\ \text{Let $F(u,v) = \int (2uv^3-3v)du$}\\ \implies F(u,v) = u^2v^3-3uv + h(v)\\ \text{Differentiating and equating it to M, we have}\\ 3u^2v^2-3u+h'(x) = 3u^2v^2-3u+4v\\ \implies h'(v) = 4v\\ h(v) = 2v^2\\ \therefore F(u,v) =u^2v^3-3uv+ 2v^2\\ \text{The solution to the exact differential equation is}\\u^2v^3-3uv+ 2v^2=c\\ \text{When u = 1, v=1, c = 0, therefore the particular solution is}\\ u^2v^3-3uv+ 2v^2=0\\ 4. (1+y^2+xy^2)dx + (x^2y+y+2xy)dy= 0\\ \text{Let $M = 1+y^2+xy^2 $ and $N = x^2y+y+2xy$}\\ \text{Let $F(x,y) = \int x^2y+y+2xydy$}\\ \implies F(x,y) = \frac{x^2y^2}{2}+\frac{y^2}{2}+xy^2 + h(x)\\ \text{Differentiating and equating it to M, we have}\\ y^2+xy^2+h'(x) = 1+y^2+xy^2\\ \implies h'(x) = 1\\ h(x) = x\\ \therefore F(x,y) =\frac{x^2y^2}{2}+\frac{y^2}{2}+xy^2 +x\\ \text{The solution to the exact differential equation is}\\-x^2y+xy^2-2y+ 3x^2=c\\ 5. (w^3+wz^2-z)dw + (z^3+w^2z-w)dz= 0\\ \text{Let $M = w^3+wz^2-z $ and $N = z^3+w^2z-w$}\\ \text{Let $F(w,z) = \int w^3+wz^2-zdw$}\\ \implies F(w,z) = \frac{w^4}{4}+\frac{w^2z^2}{2}-wz+h(z)\\ \text{Differentiating and equating it to M, we have}\\ w^2z-w+h'(x) = z^3+w^2z-w\\ \implies h'(x) = z^3\\ h(z) = \frac{z^4}{4}\\ \therefore F(x,y) =\frac{w^4}{4}+\frac{w^2z^2}{2}-wz+\frac{z^4}{4}\\ \text{The solution to the exact differential equation is}\\\frac{w^4}{4}+\frac{w^2z^2}{2}-wz+\frac{z^4}{4}=c\\ \text{When w= 4, z=2, c = 92, therefore the particular solution is}\\ \frac{w^4}{4}+\frac{w^2z^2}{2}-wz+\frac{z^4}{4}=92$