Answer to Question #276056 in Differential Equations for mark

Question

Reducible to homogeneous differential equations or by simple substitution.

1. (2𝑥 − 3𝑦 − 4)𝑑𝑥 − (3𝑥 − 4𝑦 − 2)𝑑𝑦 = 0

2. (2𝑥 − 𝑦 − 3)𝑑𝑥 − (𝑥 + 4𝑦 + 3)𝑑𝑦 = 0

3. (𝑥 − 𝑦 − 6)𝑑𝑦 = (𝑥 − 𝑦 + 2)𝑑𝑥

4. (𝑥 − 2𝑦 + 4)𝑑𝑥 + (2𝑥 − 4𝑦 − 1)𝑑𝑦 = 0

5. (𝑥 + 4𝑦 + 3)𝑑𝑥 = (2𝑥 − 𝑦 − 3)𝑑𝑦

B. Exact differential equations (include checking for exactness).

1. (𝑤3 + 𝑤𝑧 2 − 𝑧)𝑑𝑤 + (𝑧 2 + 𝑤2 𝑧 − 𝑤)𝑑𝑧 = 0

2. (cos 2𝑦 − 3𝑥 2𝑦 2 )𝑑𝑥 + (cos 2𝑦 − 2𝑥 sin 2𝑦 − 2𝑥 3𝑦)𝑑𝑦 = 0

Answer 1

A.

1.

$x\to X+h, y\to Y+k$

$2h-3k-4=0$

$3h-4k-2=0$

$h=k-2$

$2k-4-3k-4=0$

$k=-8,h=-10$

$x=X-10,y=Y-8$

$\frac{dY}{dX}=\frac{2X-3Y}{3X-4Y}$

$Y=tX,Y'=t'X+t$

$t'X+t=\frac{2X-3tX}{3X-4tX}$

$t'X=\frac{2-3t-3t+4t^2}{3-4t}$

$\frac{3-4t}{4t^2-6t+2}dt=\frac{dX}{X}$

$-\frac{ln(2t^2-3t+1)}{2}=lnX+lnc_1$

$\frac{1}{2t^2-3t+1}=c_2X^2$

$\frac{1}{2(Y/X)^2-3(Y/X)+1}=c_2X^2$

$2(\frac{y+8}{x+10})^2-3\frac{y+8}{x+10}+1=\frac{c}{(x+10)^2}$

2.

$x\to X+h, y\to Y+k$

$2h-k-3=0$

$h+4k+3=0$

$h=5k+6$

$9k+9=0$

$k=-1,h=1$

$x=X+1,y=Y-1$

$\frac{dY}{dX}=\frac{2X-Y}{X+4Y}$

$Y=tX,Y'=t'X+t$

$t'X+t=\frac{2X-tX}{X+4tX}=\frac{2-t}{1+4t}$

$t'X=\frac{-4t^2-2t+2}{1+4t}$

$\frac{1+4t}{-4t^2-2t+2}dt=\frac{dX}{X}$

$-\frac{ln(2t^2+t-1)}{2}=lnX+lnc_1$

$\frac{1}{2t^2+t-1}=c_2X^2$

$\frac{1}{2(Y/X)^2+(Y/X)-1}=c_2X^2$

$2(\frac{y+1}{x-1})^2+\frac{y+1}{x-1}-1=\frac{c}{(x-1)^2}$

3.

$x-y=v$

$1-\frac{dv}{dx}=\frac{v+2}{v-6}$

$-\frac{8}{v-6}=\frac{dv}{dx}$

$-8x=\frac{(v-6)^2}{2}+c$

$-8x=\frac{(x-y-6)^2}{2}+c$

4.

$x-2y=v$

$\frac{1}{2}(1-\frac{dv}{dx})=-\frac{v+4}{2v-1}$

$1+\frac{2v+8}{v-6}=\frac{dv}{dx}$

$\frac{3v+2}{v-6}=\frac{dv}{dx}$

$dx=\frac{v-6}{3v+2}dv$

$x=\frac{3v-20ln(3v+2)}{9}+c$

$x=\frac{3(x-2y)-20ln(3(x-2y)+2)}{9}+c$

5.

$x\to X+h, y\to Y+k$

$2h-k-3=0$

$h+4k+3=0$

$h=5k+6$

$9k+9=0$

$k=-1,h=1$

$x=X+1,y=Y-1$

$\frac{dY}{dX}=\frac{X+4Y}{2X-Y}$

$Y=tX,Y'=t'X+t$

$t'X+t=\frac{X+4tX}{2X-tX}=\frac{1+4t}{2-t}$

$t'X=\frac{-t^2+2t+1}{1+4t}$

$\frac{1+4t}{-t^2+2t+1}dt=\frac{dX}{X}$

$\frac{5}{2\sqrt 2}ln(\frac{t+\sqrt 2-1}{t-\sqrt 2-1})-2ln(t^2-2t-1)=lnX+lnc$

$\frac{5}{2\sqrt 2}ln(\frac{Y/X+\sqrt 2-1}{Y/X-\sqrt 2-1})-2ln((Y/X)^2-2Y/X-1)=ln(cX)$

$\frac{5}{2\sqrt 2}ln(\frac{(y+1)/(x-1)+\sqrt 2-1}{(y+1)/(x-1)-\sqrt 2-1})-2ln(((y+1)/(x-1))^2-2(y+1)/(x-1)-1)=$

$=ln(c(x-1))$

B.

1.

$(𝑤^3 + 𝑤𝑧 ^2 − 𝑧)_z=2zw-1$

$(𝑧 ^2 + 𝑤^2 𝑧 − 𝑤)_w=2zw-1$

$F=\int (𝑤^3 + 𝑤𝑧 ^2 − 𝑧)dw=w^4/4+w^2z^2/2-zw+g(z)$

$F_z=w^2z-w+g'(z)=𝑧 ^2 + 𝑤^2 𝑧 − 𝑤$

$g'(z)=𝑧 ^2$

$g(z)=\int𝑧 ^2dz=z^3/3+c$

$F=w^4/4+w^2z^2/2-zw+z^3/3+c$

$w^4/4+w^2z^2/2-zw+z^3/3+c=0$

2.

$(cos 2𝑦 − 3𝑥^ 2𝑦 ^2 )_y=-2sin2y-6x^2y$

$(cos 2𝑦 − 2𝑥 sin 2𝑦 − 2𝑥 ^3𝑦)_x=-2sin2y-6x^2y$

$F=\int (cos 2𝑦 − 3𝑥^ 2𝑦 ^2 )dx=xcos2y-x^3y^2+g(y)$

$F_y=-2xsin2y-2yx^3+g'(y)=cos 2𝑦 − 2𝑥 sin 2𝑦 − 2𝑥 ^3𝑦$

$g'(y)=cos 2𝑦$

$g(y)=\intop cos2ydy=sin2y/2+c$

$F=xcos2y-x^3y^2+sin2y/2+c$

$xcos2y-x^3y^2+sin2y/2+c=0$