## Ayers ‘Differential Equations Problem 4

Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.

4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2

Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))

Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0

When x=1,y=2: (2-C)^2 = C and C = 1,4

The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x

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The primitive is given as (y-C)^2=Cx

derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C

He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive

derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C

Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :

dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)

Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is

(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :

C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left

C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4

which is quadratic, complete the square :

C^2-5C+2.5^2 = -4+2.5^2

(C-2.5)^2 = -4+6.25

(C-2.5)^2 = 2.25 , square root each side

C-2.5 = + – SquareRoot(2.25)

C = +1.5+2.5 or -1.5+2.5

C = 4 or 1

The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0

combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0

cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0

common denominator (y-C)^2

C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0

Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0

The Cx(y-C) term in numerator can be factored Cxy-CxC

(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x

(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.

4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2

Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))

Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0

When x=1,y=2: (2-C)^2 = C and C = 1,4

The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x

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%%%%%%%%%%% My own commentary

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The primitive is given as (y-C)^2=Cx

derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C

He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive

derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C

Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :

dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)

Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is

(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :

C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left

C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4

which is quadratic, complete the square :

C^2-5C+2.5^2 = -4+2.5^2

(C-2.5)^2 = -4+6.25

(C-2.5)^2 = 2.25 , square root each side

C-2.5 = + – SquareRoot(2.25)

C = +1.5+2.5 or -1.5+2.5

C = 4 or 1

The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0

combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0

cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0

common denominator (y-C)^2

C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0

Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0

The Cx(y-C) term in numerator can be factored Cxy-CxC

(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x

(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2 Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.

4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2

Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))

Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0

When x=1,y=2: (2-C)^2 = C and C = 1,4

The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x

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%%%%%%%%%%% My own commentary

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The primitive is given as (y-C)^2=Cx

derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C

He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive

derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C

Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :

dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)

Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is

(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :

C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left

C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4

which is quadratic, complete the square :

C^2-5C+2.5^2 = -4+2.5^2

(C-2.5)^2 = -4+6.25

(C-2.5)^2 = 2.25 , square root each side

C-2.5 = + – SquareRoot(2.25)

C = +1.5+2.5 or -1.5+2.5

C = 4 or 1

The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0

combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0

cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0

common denominator (y-C)^2

C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0

Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0

The Cx(y-C) term in numerator can be factored Cxy-CxC

(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x

(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2 ayers problem4 MATLAB Answers — New Questions