Ayers 'Differential Equations Problem 4

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Joseph Palumbo
Joseph Palumbo 2024-5-18
评论： Joseph Palumbo 2024-5-24
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

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回答（2 个）

Athanasios Paraskevopoulos
Ran in:
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
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Joseph Palumbo
Joseph Palumbo 2024-5-22
I see you ran it on 2024(a) but mine is R2022(b) could this be the problem?
Joseph Palumbo
Joseph Palumbo 2024-5-22
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

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Joseph Palumbo
Joseph Palumbo 2024-5-24
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

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