Question #276056

Reducible to homogeneous differential equations or by simple substitution.

1. (2π‘₯ βˆ’ 3𝑦 βˆ’ 4)𝑑π‘₯ βˆ’ (3π‘₯ βˆ’ 4𝑦 βˆ’ 2)𝑑𝑦 = 0

2. (2π‘₯ βˆ’ 𝑦 βˆ’ 3)𝑑π‘₯ βˆ’ (π‘₯ + 4𝑦 + 3)𝑑𝑦 = 0

3. (π‘₯ βˆ’ 𝑦 βˆ’ 6)𝑑𝑦 = (π‘₯ βˆ’ 𝑦 + 2)𝑑π‘₯

4. (π‘₯ βˆ’ 2𝑦 + 4)𝑑π‘₯ + (2π‘₯ βˆ’ 4𝑦 βˆ’ 1)𝑑𝑦 = 0

5. (π‘₯ + 4𝑦 + 3)𝑑π‘₯ = (2π‘₯ βˆ’ 𝑦 βˆ’ 3)𝑑𝑦


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

1. (𝑀3 + 𝑀𝑧 2 βˆ’ 𝑧)𝑑𝑀 + (𝑧 2 + 𝑀2 𝑧 βˆ’ 𝑀)𝑑𝑧 = 0

2. (cos 2𝑦 βˆ’ 3π‘₯ 2𝑦 2 )𝑑π‘₯ + (cos 2𝑦 βˆ’ 2π‘₯ sin 2𝑦 βˆ’ 2π‘₯ 3𝑦)𝑑𝑦 = 0


1
Expert's answer
2021-12-07T09:55:59-0500

A.

1.

x→X+h,y→Y+kx\to X+h, y\to Y+k

2hβˆ’3kβˆ’4=02h-3k-4=0

3hβˆ’4kβˆ’2=03h-4k-2=0

h=kβˆ’2h=k-2

2kβˆ’4βˆ’3kβˆ’4=02k-4-3k-4=0

k=βˆ’8,h=βˆ’10k=-8,h=-10

x=Xβˆ’10,y=Yβˆ’8x=X-10,y=Y-8


dYdX=2Xβˆ’3Y3Xβˆ’4Y\frac{dY}{dX}=\frac{2X-3Y}{3X-4Y}


Y=tX,Yβ€²=tβ€²X+tY=tX,Y'=t'X+t


tβ€²X+t=2Xβˆ’3tX3Xβˆ’4tXt'X+t=\frac{2X-3tX}{3X-4tX}


tβ€²X=2βˆ’3tβˆ’3t+4t23βˆ’4tt'X=\frac{2-3t-3t+4t^2}{3-4t}


3βˆ’4t4t2βˆ’6t+2dt=dXX\frac{3-4t}{4t^2-6t+2}dt=\frac{dX}{X}


βˆ’ln(2t2βˆ’3t+1)2=lnX+lnc1-\frac{ln(2t^2-3t+1)}{2}=lnX+lnc_1


12t2βˆ’3t+1=c2X2\frac{1}{2t^2-3t+1}=c_2X^2


12(Y/X)2βˆ’3(Y/X)+1=c2X2\frac{1}{2(Y/X)^2-3(Y/X)+1}=c_2X^2


2(y+8x+10)2βˆ’3y+8x+10+1=c(x+10)22(\frac{y+8}{x+10})^2-3\frac{y+8}{x+10}+1=\frac{c}{(x+10)^2}


2.

x→X+h,y→Y+kx\to X+h, y\to Y+k

2hβˆ’kβˆ’3=02h-k-3=0

h+4k+3=0h+4k+3=0

h=5k+6h=5k+6

9k+9=09k+9=0

k=βˆ’1,h=1k=-1,h=1

x=X+1,y=Yβˆ’1x=X+1,y=Y-1


dYdX=2Xβˆ’YX+4Y\frac{dY}{dX}=\frac{2X-Y}{X+4Y}


Y=tX,Yβ€²=tβ€²X+tY=tX,Y'=t'X+t


tβ€²X+t=2Xβˆ’tXX+4tX=2βˆ’t1+4tt'X+t=\frac{2X-tX}{X+4tX}=\frac{2-t}{1+4t}


tβ€²X=βˆ’4t2βˆ’2t+21+4tt'X=\frac{-4t^2-2t+2}{1+4t}


1+4tβˆ’4t2βˆ’2t+2dt=dXX\frac{1+4t}{-4t^2-2t+2}dt=\frac{dX}{X}


βˆ’ln(2t2+tβˆ’1)2=lnX+lnc1-\frac{ln(2t^2+t-1)}{2}=lnX+lnc_1


12t2+tβˆ’1=c2X2\frac{1}{2t^2+t-1}=c_2X^2


12(Y/X)2+(Y/X)βˆ’1=c2X2\frac{1}{2(Y/X)^2+(Y/X)-1}=c_2X^2


2(y+1xβˆ’1)2+y+1xβˆ’1βˆ’1=c(xβˆ’1)22(\frac{y+1}{x-1})^2+\frac{y+1}{x-1}-1=\frac{c}{(x-1)^2}


3.

xβˆ’y=vx-y=v


1βˆ’dvdx=v+2vβˆ’61-\frac{dv}{dx}=\frac{v+2}{v-6}


βˆ’8vβˆ’6=dvdx-\frac{8}{v-6}=\frac{dv}{dx}


βˆ’8x=(vβˆ’6)22+c-8x=\frac{(v-6)^2}{2}+c


βˆ’8x=(xβˆ’yβˆ’6)22+c-8x=\frac{(x-y-6)^2}{2}+c


4.

xβˆ’2y=vx-2y=v


12(1βˆ’dvdx)=βˆ’v+42vβˆ’1\frac{1}{2}(1-\frac{dv}{dx})=-\frac{v+4}{2v-1}


1+2v+8vβˆ’6=dvdx1+\frac{2v+8}{v-6}=\frac{dv}{dx}


3v+2vβˆ’6=dvdx\frac{3v+2}{v-6}=\frac{dv}{dx}


dx=vβˆ’63v+2dvdx=\frac{v-6}{3v+2}dv


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


x=3(xβˆ’2y)βˆ’20ln(3(xβˆ’2y)+2)9+cx=\frac{3(x-2y)-20ln(3(x-2y)+2)}{9}+c


5.

x→X+h,y→Y+kx\to X+h, y\to Y+k

2hβˆ’kβˆ’3=02h-k-3=0

h+4k+3=0h+4k+3=0


h=5k+6h=5k+6

9k+9=09k+9=0

k=βˆ’1,h=1k=-1,h=1


x=X+1,y=Yβˆ’1x=X+1,y=Y-1


dYdX=X+4Y2Xβˆ’Y\frac{dY}{dX}=\frac{X+4Y}{2X-Y}


Y=tX,Yβ€²=tβ€²X+tY=tX,Y'=t'X+t


tβ€²X+t=X+4tX2Xβˆ’tX=1+4t2βˆ’tt'X+t=\frac{X+4tX}{2X-tX}=\frac{1+4t}{2-t}


tβ€²X=βˆ’t2+2t+11+4tt'X=\frac{-t^2+2t+1}{1+4t}


1+4tβˆ’t2+2t+1dt=dXX\frac{1+4t}{-t^2+2t+1}dt=\frac{dX}{X}


522ln(t+2βˆ’1tβˆ’2βˆ’1)βˆ’2ln(t2βˆ’2tβˆ’1)=lnX+lnc\frac{5}{2\sqrt 2}ln(\frac{t+\sqrt 2-1}{t-\sqrt 2-1})-2ln(t^2-2t-1)=lnX+lnc


522ln(Y/X+2βˆ’1Y/Xβˆ’2βˆ’1)βˆ’2ln((Y/X)2βˆ’2Y/Xβˆ’1)=ln(cX)\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)


522ln((y+1)/(xβˆ’1)+2βˆ’1(y+1)/(xβˆ’1)βˆ’2βˆ’1)βˆ’2ln(((y+1)/(xβˆ’1))2βˆ’2(y+1)/(xβˆ’1)βˆ’1)=\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(xβˆ’1))=ln(c(x-1))


B.

1.

(𝑀3+𝑀𝑧2βˆ’π‘§)z=2zwβˆ’1(𝑀^3 + 𝑀𝑧 ^2 βˆ’ 𝑧)_z=2zw-1

(𝑧2+𝑀2π‘§βˆ’π‘€)w=2zwβˆ’1(𝑧 ^2 + 𝑀^2 𝑧 βˆ’ 𝑀)_w=2zw-1


F=∫(𝑀3+𝑀𝑧2βˆ’π‘§)dw=w4/4+w2z2/2βˆ’zw+g(z)F=\int (𝑀^3 + 𝑀𝑧 ^2 βˆ’ 𝑧)dw=w^4/4+w^2z^2/2-zw+g(z)


Fz=w2zβˆ’w+gβ€²(z)=𝑧2+𝑀2π‘§βˆ’π‘€F_z=w^2z-w+g'(z)=𝑧 ^2 + 𝑀^2 𝑧 βˆ’ 𝑀

gβ€²(z)=𝑧2g'(z)=𝑧 ^2

g(z)=βˆ«π‘§2dz=z3/3+cg(z)=\int𝑧 ^2dz=z^3/3+c


F=w4/4+w2z2/2βˆ’zw+z3/3+cF=w^4/4+w^2z^2/2-zw+z^3/3+c

w4/4+w2z2/2βˆ’zw+z3/3+c=0w^4/4+w^2z^2/2-zw+z^3/3+c=0


2.

(cos2π‘¦βˆ’3π‘₯2𝑦2)y=βˆ’2sin2yβˆ’6x2y(cos 2𝑦 βˆ’ 3π‘₯^ 2𝑦 ^2 )_y=-2sin2y-6x^2y

(cos2π‘¦βˆ’2π‘₯sin2π‘¦βˆ’2π‘₯3𝑦)x=βˆ’2sin2yβˆ’6x2y(cos 2𝑦 βˆ’ 2π‘₯ sin 2𝑦 βˆ’ 2π‘₯ ^3𝑦)_x=-2sin2y-6x^2y


F=∫(cos2π‘¦βˆ’3π‘₯2𝑦2)dx=xcos2yβˆ’x3y2+g(y)F=\int (cos 2𝑦 βˆ’ 3π‘₯^ 2𝑦 ^2 )dx=xcos2y-x^3y^2+g(y)


Fy=βˆ’2xsin2yβˆ’2yx3+gβ€²(y)=cos2π‘¦βˆ’2π‘₯sin2π‘¦βˆ’2π‘₯3𝑦F_y=-2xsin2y-2yx^3+g'(y)=cos 2𝑦 βˆ’ 2π‘₯ sin 2𝑦 βˆ’ 2π‘₯ ^3𝑦

gβ€²(y)=cos2𝑦g'(y)=cos 2𝑦


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


F=xcos2yβˆ’x3y2+sin2y/2+cF=xcos2y-x^3y^2+sin2y/2+c

xcos2yβˆ’x3y2+sin2y/2+c=0xcos2y-x^3y^2+sin2y/2+c=0


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