Solve the differential equation (dy/dx) = (2y - x + 7) / (4x - 3y - 18)?
Have to use the method of finding h and k so that the substitutions x=u+h, y=v+k transform the differential equation (dy/dx)=(x-y-1)/(x+y+3) into the homogeneous equation (dv/du)=(u-v)/(u+v).
I know I have to make two substitutions to solve it. I set u-v=2y-x+7 and u+v=4x-3y-18 (right?) and solve for u and v but I'm not getting anywhere when trying to find du and dv. Help.
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Verified answer
(dy/dx) = (2y - x + 7) / (4x - 3y - 18)
(4x - 3y - 18)dy = (2y - x + 7)dx
(4x - 3y - 18)dy - (2y - x + 7)dx = 0
x = u+h ===> dx/du = 1 or dx = du
y = v+k ===> dy/dv = 1 or dy = dv
(4x - 3y - 18)dy - (2y - x + 7)dx = 0
(4u + 4h - 3v - 3k - 18)dy - (2v + 2k - u - h + 7)dx = 0
but dx = du and dy = dv; so
(4u - 3v + 4h - 3k - 18)dv - (2v - u + 2k - h + 7)du = 0
solve for h and k
4h - 3k - 18 = 0 and 2k - h + 7 = 0
4h - 3k = 18 and 2k - h = -7
h = 3, k = -2
x = u+h ==> x = u + 3
y = v+k ==> y = v - 2
(4u - 3v)dv - (2v - u )du = 0
(4u - 3v)dv = (2v - u )du
dv/du = (2v - u ) /(4u - 3v)
v' = (2v - u ) /(4u - 3v)
let v = uw and differentiate with respect to u
v' = uw' + w
equate 'em
uw' + w = (2v - u ) /(4u - 3v)
uw' + w = (2uw - u ) /(4u - 3uw)
uw' = (2uw - u ) /(4u - 3uw) - w
uw' = (2uw - u - 4uw + 3uw²) /(4u - 3uw)
factor out u on the right-hand side of the equation to get
uw' = (3w² - 2w - 1) /(4 - 3w)
dw/du = (3w² - 2w - 1) /(4 - 3w)
[ (4 - 3w) /(3w² - 2w - 1) ] dw = du/u
integrate both sides
let z = 3w² - 2w - 1
z' = 6w - 2
∫ [ (4 - 3w) /(3w² - 2w - 1) ] dw = ∫ du/u
[ (4 - 3w) /(3w² - 2w - 1) ] dw = du/u
1/4 ln(9-9w) - 5/4 ln(9w+3) = lnu + lnC
ln(9-9w) - 5ln(9w+3) = 4lnu + lnA
ln[ (9-9w) / (9w+3)^5 ] = ln(Au^4)
(9-9w) / (9w+3)^5 ] = Au^4
but v = uw ; so w = v/u
replace w with v/u and also replace v and u with values from x = u+h ==> x = u + 3
y = v+k ==> y = v - 2
and should end up with
(9x - 9y - 45) / (9y + 3x + 9)^5 = A
9(x - y - 5) / [ 243(3y + 3x + 3)^5 ] = A
(x - y - 5) / (3y + 3x + 3)^5 = C
(x - y - 5) = C [ (3y + 3x + 3)^5 ]
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where C = 27A
hmmm ... 8x - 2y = 33 ... 4x +3y = 9 (multply this eqn via 2) to get: 8x + 6y = 18 .... subtract it fromthe first eqn -8y = 15 y = -15/8 .... now subst this cost into first eqn 8x - 2(-15/8) = 33 8x + 15/4 = 33 8x = 33 - 15/4 = 117 / 4 => x = 117 / 32 soooooo x = 117 / 32 y = -15 / 8 try answer: in 8x - 27 = 33 8*117/32 - 2*(-15/8) = 117/4 + 15/4 = 132 / 4 = 33 seems ok