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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%%%%%%%%%%% 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

Answer 1

[Athanasios Paraskevopoulos]

syms x y(x) C

% Define the differential equation
dy_dx = diff(y, x);
ode = 4*x*dy_dx^2 + 2*x*dy_dx - y == 0;

% Solve the differential equation
sol = dsolve(ode);

% Display the general solution
disp('General Solution:');

General Solution:

disp(sol);

% Verify that (y - C)^2 = Cx is the primitive
eq_primitive = (y - C)^2 == C*x;

% Substitute the general solution into the primitive equation
sol_y1 = sol(1);
sol_y2 = sol(2);

disp('Verifying the primitive for first solution:');

Verifying the primitive for first solution:

simplify(subs(eq_primitive, y, sol_y1))

ans(x) = 

disp('Verifying the primitive for second solution:');

Verifying the primitive for second solution:

simplify(subs(eq_primitive, y, sol_y2))

ans(x) = 

% Find the equations of the integral curves through the points (x = 1, y = 2)
x_val = 1;
y_val = 2;

C1 = solve(subs((y - C)^2, {x, y}, {x_val, y_val}) == C*x_val, C);

disp('Values of C:');

Values of C:

disp(C1);

% Integral curves equations
disp('Integral curves equations:');

Integral curves equations:

for i = 1:length(C1)
    eq_curve = (y - C1(i))^2 == C1(i)*x;
    disp(eq_curve);
end

[Joseph Palumbo]

I do not get 4C-2Sqrt(C1*x),4C+2Sqrt(C1*x),-x/4 and 0 for the general solution, rather I get 0 and 1/2+x/4, consequently I get

C == x|C == 0 for primitive of the 1st solution and (2 - 4*C + x)^2 == 16*C*x for primitive of 2nd solution

Here is my code:

syms x y(x) C

% Define the differential equation

dy_dx = diff(y, x);

ode = 4*x*dy_dx^2 + 2*x*dy_dx - y == 0;

% Solve the differential equation

sol = dsolve(ode);

% Verify that (y-C)^2=Cx is the primitive

eq_primitive=(y-C)^2==C*x;

% Substitute the general solution into the primitive equation

sol_y1=sol(1);

sol_y2=sol(2);

disp("Verifying the primitive for the first solution:");

simplify(subs(eq_primitive,y,sol_y1))

disp("Verifying the primitive for the second solution:");

simplify(subs(eq_primitive,y,sol_y2))

% C=xvC=0 output is really C == x|C == 0, I do not know what this means

% perhaps element wise 'or'

% Find the equations of the integral curves through the points x=1 y=2

x_val=1;

y_val=2;

C1=solve(subs((y-C)^2,{x,y},{x_val,y_val})==C*x_val,C);

disp("Values of C");

disp(C1);

% Integral Curves Equations

disp("Integral curves equations:");

for i=1:length(C1)

eq_curve=(y-C1(i))^2==C1(i)*x;

disp(eq_curve);

end

Here is my output

diff(y(x), x)

General Solution:

[sym(0); sym(1/2) + x/4]

Verifying the primitive for the first solution:

C == x|C == 0

Verifying the primitive for the second solution:

(2 - 4*C + x)^2 == 16*C*x

Values of C

[sym(1); sym(4)]

Integral curves equations:

(y(x) - 1)^2 == x

(y(x) - 4)^2 == 4*x

Everything else seems to be correct Thanks alot for helping me

[Joseph Palumbo]

Sorry , I did not display the first few lines but I just added them in it's the same as you advised

[Joseph Palumbo]

I see you ran it on 2024(a) but mine is R2022(b) could this be the problem?

[Joseph Palumbo]

For some reason my earlier reply was not recorded:

I do not get 4C-2Sqrt(C1*x),4C+2Sqrt(C1*x),-x/4 and 0 for the general solution, rather I get 0 and 1/2+x/4, consequently I get

C == x|C == 0 for primitive of the 1st solution and (2 - 4*C + x)^2 == 16*C*x for primitive of 2nd solution

Here is my code:

% Verify that (y-C)^2=Cx is the primitive

eq_primitive=(y-C)^2==C*x;

% Substitute the general solution into the primitive equation

sol_y1=sol(1);

sol_y2=sol(2);

disp("Verifying the primitive for the first solution:");

simplify(subs(eq_primitive,y,sol_y1))

disp("Verifying the primitive for the second solution:");

simplify(subs(eq_primitive,y,sol_y2))

% C=xvC=0 output is really C == x|C == 0, I do not know what this means

% perhaps element wise 'or'

% Find the equations of the integral curves through the points x=1 y=2

x_val=1;

y_val=2;

C1=solve(subs((y-C)^2,{x,y},{x_val,y_val})==C*x_val,C);

disp("Values of C");

disp(C1);

% Integral Curves Equations

disp("Integral curves equations:");

for i=1:length(C1)

eq_curve=(y-C1(i))^2==C1(i)*x;

disp(eq_curve);

end

Here is my output

diff(y(x), x)

General Solution:

[sym(0); sym(1/2) + x/4]

Verifying the primitive for the first solution:

C == x|C == 0

Verifying the primitive for the second solution:

(2 - 4*C + x)^2 == 16*C*x

Values of C

[sym(1); sym(4)]

Integral curves equations:

(y(x) - 1)^2 == x

(y(x) - 4)^2 == 4*x

Everything else seems to be correct Thanks alot for helping me

Answer 2

[Joseph Palumbo]

I think this might be it, hope I just didn't understand what given by others to help me, but possibly the problem is just an exercise in Mathematics not a real application problem, anyway I believe this is it:

clearvars

syms x y C

disp('The primitive was given as (y-C)^2=Cx')

disp("Justified by laws of Mahtematics to differentiate ")

disp("each term of the primitive and place the unknown differential dy/dx")

disp("to be solved for in as a factor thus diff((y-C)^2)*dy/dx=diff(Cx)")

disp("and solve algebraically")

% differentiate Cx answer to primitive

disp("diff(Cx)=")

D1=diff(C*x)

% differentiate (y-C)^2 of the primitive

disp("diff(y-C)^2)=")

D2=diff((y-C)^2)

disp("Now placing unknown dy/dx as a factor and solving for dy/dx")

% Solve the differential equation

D=D1/D2;

% display the general solution

disp("General Solution:");

D

% Prove the primitive

disp("If dy/dx = diff(Cx)/diff((y-C)^2) then the primitive which was (y-C)^2")

%disp("=Integrate(C/C/2(y-C)")

%D=int(2*(y-C),[y C])

% making the denominator negative changes signs thus -2C+2y which factors

% to 2(y-C) thus matching Ayres and the book : dy/dx=C/2(y-C)

% making the denominator negative changes signs thus -2C+2y which factors

% to 2(y-C) thus matching Ayres and the book : dy/dx=C/2(y-C)")

disp("if result is negative either numerator or denominator can be negative" )

disp("but not both making the denominator negative changes signs");

disp("to 2(y-C) thus matching Ayres and the book : dy/dx=C/2(y-C)")

% same being true here where -(C-y)^2 is actually (y-C)^2

disp("same being true here, where -(C-y)^2 is actually (y-C)^2")

D=D1/D2

Dcopy=D;

D=C/(2*(y-C)^2)

% Prove the primitive

disp("If dy/dx = diff(Cx)/diff((y-C)^2) then the primitive which was (y-C)^2")

disp("=Integrate(C/C/2(y-C)^2")

D=int(D1/Dcopy,[y C])

disp("Again negative changes signs within parenthesis thus -(C-y)^2=(y-C)^2")

prmitive=(y-C)^2

% Find the equations of the integral curves through the points x=1 y=2

x_val=1;

y_val=2;

C1=solve(subs((y-C)^2,{x,y},{x_val,y_val})==C*x_val,C);

disp("Values of C");

disp(C1);

% Integral Curves Equations

disp("Integral curves equations:");

for i=1:length(C1)

eq_curve=(y-C1(i))^2==C1(i)*x;

disp(eq_curve);

end

[Joseph Palumbo]

This post is closed Joseph Palumbo