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Pure Maths Questions , please help !
i got 4 questions stuck
find the general solution of each differential equation using the subsititustion given where v is a function of x.
1) dy/dx= y(x+2y) / x(y+2x), x <> 0 ; y=vx
2)(x+y)dy/dx = x*x + xy+x+1 ; y=v-x
3)x d^2/dx^2- 2 dy/dx + x= 0 ; dy/dx = v
4) given that y=1 at x=2, use the substitustion v=3x-y-3 to solve the differential equation
(3x-y-1) dy/dx = (3x-y+3)
Please help , ThX
find the general solution of each differential equation using the subsititustion given where v is a function of x.
1) dy/dx= y(x+2y) / x(y+2x), x <> 0 ; y=vx
2)(x+y)dy/dx = x*x + xy+x+1 ; y=v-x
3)x d^2/dx^2- 2 dy/dx + x= 0 ; dy/dx = v
4) given that y=1 at x=2, use the substitustion v=3x-y-3 to solve the differential equation
(3x-y-1) dy/dx = (3x-y+3)
Please help , ThX
1)
y=vx => dy/dx = v + x dv/dx
dy/dx= y(x+2y) / x(y+2x),
=> v + x dv/dx = vx(x+2vx)/ x(vx+2x)
= (vx² + 2v²x² )/(vx² + 2x²)
=> x dv/dx = (vx² + 2v²x² - v²x² + 2vx²)/(vx² + 2x²)
=> dv/dx = (3v + v²)/(vx + 2x)
=> (v+2)/v(3+v) dv/dx = 1/x
=> (v+3/2)/v(3+v) - 1/2v(3+v) dv/dx= 1/x
=> (v+3/2)/v(3+v) - 1/6v + 1/6(3+v) dv/dx = 1/x
=> 1/2 ln|v(3+v)| - 1/6 ln|v| + 1/6 ln|3+v| = ln|Ax|
=> ln|(3+v)^(3/2)/(v)^(1/2)| = ln |Ax|
=> (3+v)^3/2 / v^1/2 = Ax
v = y/x => (3+y/x)^3/2 / (y/x)^1/2 = Ax
=> (3+y/x)³ = Ax²(y/x) = Ayx
27 + 27y/x + 9y²/x² + y³/x³ = Ayx
27x³ + 27yx² + 9y²x + y³ = Ayx^4
=> Ayx^4 - 27x³ - 27yx² - 9y²x - y³ = 0
y=vx => dy/dx = v + x dv/dx
dy/dx= y(x+2y) / x(y+2x),
=> v + x dv/dx = vx(x+2vx)/ x(vx+2x)
= (vx² + 2v²x² )/(vx² + 2x²)
=> x dv/dx = (vx² + 2v²x² - v²x² + 2vx²)/(vx² + 2x²)
=> dv/dx = (3v + v²)/(vx + 2x)
=> (v+2)/v(3+v) dv/dx = 1/x
=> (v+3/2)/v(3+v) - 1/2v(3+v) dv/dx= 1/x
=> (v+3/2)/v(3+v) - 1/6v + 1/6(3+v) dv/dx = 1/x
=> 1/2 ln|v(3+v)| - 1/6 ln|v| + 1/6 ln|3+v| = ln|Ax|
=> ln|(3+v)^(3/2)/(v)^(1/2)| = ln |Ax|
=> (3+v)^3/2 / v^1/2 = Ax
v = y/x => (3+y/x)^3/2 / (y/x)^1/2 = Ax
=> (3+y/x)³ = Ax²(y/x) = Ayx
27 + 27y/x + 9y²/x² + y³/x³ = Ayx
27x³ + 27yx² + 9y²x + y³ = Ayx^4
=> Ayx^4 - 27x³ - 27yx² - 9y²x - y³ = 0
elpaw
1)
y=vx => dy/dx = v + x dv/dx
dy/dx= y(x+2y) / x(y+2x),
=> v + x dv/dx = vx(x+2vx)/ x(vx+2x)
= (vx² + 2v²x² )/(vx² + 2x²)
=> x dv/dx = (vx² + 2v²x² - v²x² + 2vx²)/(vx² + 2x²)
=> dv/dx = (3v + v²)/(vx + 2x)
=> (v+2)/v(3+v) dv/dx = 1/x
=> (v+3/2)/v(3+v) - 1/2v(3+v) dv/dx= 1/x
=> (v+3/2)/v(3+v) - 1/6v + 1/6(3+v) dv/dx = 1/x
=> 1/2 ln|v(3+v)| - 1/6 ln|v| + 1/6 ln|3+v| = ln|Ax|
=> ln|(3+v)^(3/2)/(v)^(1/2)| = ln |Ax|
=> (3+v)^3/2 / v^1/2 = Ax
v = y/x => (3+y/x)^3/2 / (y/x)^1/2 = Ax
=> (3+y/x)³ = Ax²(y/x) = Ayx
27 + 27y/x + 9y²/x² + y³/x³ = Ayx
27x³ + 27yx² + 9y²x + y³ = Ayx^4
=> Ayx^4 - 27x³ - 27yx² - 9y²x - y³ = 0
y=vx => dy/dx = v + x dv/dx
dy/dx= y(x+2y) / x(y+2x),
=> v + x dv/dx = vx(x+2vx)/ x(vx+2x)
= (vx² + 2v²x² )/(vx² + 2x²)
=> x dv/dx = (vx² + 2v²x² - v²x² + 2vx²)/(vx² + 2x²)
=> dv/dx = (3v + v²)/(vx + 2x)
=> (v+2)/v(3+v) dv/dx = 1/x
=> (v+3/2)/v(3+v) - 1/2v(3+v) dv/dx= 1/x
=> (v+3/2)/v(3+v) - 1/6v + 1/6(3+v) dv/dx = 1/x
=> 1/2 ln|v(3+v)| - 1/6 ln|v| + 1/6 ln|3+v| = ln|Ax|
=> ln|(3+v)^(3/2)/(v)^(1/2)| = ln |Ax|
=> (3+v)^3/2 / v^1/2 = Ax
v = y/x => (3+y/x)^3/2 / (y/x)^1/2 = Ax
=> (3+y/x)³ = Ax²(y/x) = Ayx
27 + 27y/x + 9y²/x² + y³/x³ = Ayx
27x³ + 27yx² + 9y²x + y³ = Ayx^4
=> Ayx^4 - 27x³ - 27yx² - 9y²x - y³ = 0
how long did that take you to work out, u mus b clever
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