Differential Equation Ayers Chapter 2 problem 4

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

[Joseph Palumbo]

How do I solve the general solution, prove the primitive

[Joseph Palumbo]

The problem was from Ayers Differential equations Schaums Outline Chap. 2 problem 4

Show that (y-C)^2=Cx is the primitive of the differential equation 4x*(dx/dy)^2 + 2xdy/dx – y=0

And find the equations of the integral curves through the point (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)62=0

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

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

Athanasios Paraskevopoulos

Graciously helped me with the last parts of it, but the general solution I cannot get the same thing as he does and proving the primitive I do not get the same as he does, he used R2024(A) I’m using R2022(b),

But I do not understand the answers he gets, I showed the code and the output I get.

I want to thank everybody for even reading this and trying to help---- Joseph Palumbo

Answer 1

[Arnav]

Hi @Joseph Palumbo

You outlined 2 tasks in the comments:

• Showing that is the primitive of the differential equation

This can be done in the following steps:

    syms x y C;
    x = solve((y - C)^2 == C*x, x);
    dy_dx = 1 / diff(x,y);
    diff_eq = 4*x*(dy_dx)^2 + 2*x*dy_dx - y;
    LHS = simplify(diff_eq) %Verify that LHS simplifies to 0 i.e. RHS

LHS = 

0

It can be verified that the LHS simplifies to 0. Hence, the given curve is the primitive of the differential equation.

• Finding particular solutions that pass through the point .

This is handled in the code you provided. In particular, the solve statement:

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

is calculating the values of C that satisfies the equation after the values of x and y are substituted into the primitive curve.

To learn more about how solve function works, you may refer here:

https://www.mathworks.com/help/releases/R2022b/symbolic/sym.solve.html