solving the differential equation

2021 12 15
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更新时间：2021 12 15
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Hi.
I have a differential equation of a function f(x,y) wrt "x". I can solve this equation using the "dsolve" and obtain the solution. Now I again want to differentiate the solution obtained wrt to other variable "y". Can anyone please help me in writing the code for differentiating the solution thus obtained. Here is the code attached:
syms f(x) y
ode = diff(f,x) == -(2/x)*f + (4*x^3*y^2 + 2*x*y + 6*x)/x^2;
sol = dsolve(ode);
%-------------------------------------------------%
val =
C3/x^2 + (x^2*(y + 3) + x^4*y^2)/x^2
In this code, I want to differentiate the "sol" wrt variable "y". Note that the constant "C3" in the solution is a function of "x & y".

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Chunru
Chunru 2021-12-15
If C3 is an arbitrary function of y, then the differentiation wrt y can be also arbitrary. So you have to determine what C3 is (perhaps with other conditions) before you can do the differentiation to y.
mukesh bisht
mukesh bisht 2021-12-15
编辑：Walter Roberson 2021-12-15
Hey, the only difficulty I am having is writing the code (perhaps one line) so that I obtain the differential equation when I differentiate the "sol" wrt variable "y".
If I use the function "diff" to differentiate "sol" I get the following result:
ode_2 = diff(sol,y)
ans : ode_2 =
(2*y*x^4 + x^2)/x^2
Here it treats "C3" as a constant and therefore on differentiating, the whole term "C3/x^2" becomes 0.
mukesh bisht
mukesh bisht 2021-12-15
Alternatively, can you help me to solve this PDE in matlab
Boundary condition:
Walter Roberson
Walter Roberson 2021-12-15
Ran in:
Ran in:
syms f(x,y)
ode = diff(f,x) + diff(f,y) + (2/x)*f == (4*x^3*y^2 + 2*x*y + 6*x)/x^2 + 2*x^2*y + 1
ode(x, y) = 
Walter Roberson
Walter Roberson 2021-12-15
Ran in:
Ran in:
syms f(x) y C(y)
ode = diff(f,x) == -(2/x)*f + (4*x^3*y^2 + 2*x*y + 6*x)/x^2;
sol = simplify(expand((dsolve(ode))))
sol = 
vars = symvar(sol);
soly = subs(sol, vars(1), C)
soly = 
Dsoly = diff(soly,y)
Dsoly = 

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