# differential equation with mixed linear and log derivatives - proper setting

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PatrizioGraziosi on 17 Jun 2020
Commented: Ameer Hamza on 17 Jun 2020
Hello everybody,
I'd like to solve for y = y(x) the following equation
that contains derivatives on both x and log(x).
When I input the equation as
syms y(x) f
eq = diff( log(y), log(x) ) + diff( log(diff(y,x)), log(x) ) + diff( y, log(x) )*log(x) + y ...
== 1 + f;
Second argument must be a variable or a nonnegative integer specifying the number of
differentiations.
I have tried to input it as a system of equations
syms y(x,z) f
eq1 = diff( log(y), x ) + diff( log(diff(y,z)), x ) + diff( y, x )*x + y ...
== 1 + f;
eq2 = x == log(z);
But when I try to solve it
odes = [eq1;eq2];
sol = dsolve(odes);
I get an error that
Symbolic ODEs must have exactly one independent variable.
I'm likely doing something wrong in managing the equations.
Thanks,
Patrizio

Ameer Hamza on 17 Jun 2020
Edited: Ameer Hamza on 17 Jun 2020
Using chain-rule, we can write
Therefore, the equation can be written as
syms y(x) f
eq = diff(log(y),x)*1/diff(log(x),x) + diff(log(diff(y,x)),x)*1/diff(log(x),x) + ...
diff(y,x)*1/diff(log(x),x)*log(x) + y ...
== 1 + f;
sol = dsolve(eq);
The symbolic solution is
>> sol
sol =
((2*C2*x^y + C2*f*x^y - C2*x^y*y + 2*C1*x^f*x^2)/(x^y*(f - y + 2)))^(1/2)
-((2*C2*x^y + C2*f*x^y - C2*x^y*y + 2*C1*x^f*x^2)/(x^y*(f - y + 2)))^(1/2)
For numerical solution, try this
syms y(x) f
eq = diff(log(y),x)*1/diff(log(x),x) + diff(log(diff(y,x)),x)*1/diff(log(x),x) + ...
diff(y,x)*1/diff(log(x),x)*log(x) + y ...
== 1 + f;
eq2 = odeToVectorField(eq);
odeFun = matlabFunction(eq2, 'Vars', {'x', 'Y', 'f'});
xspan = [0.1 10];
xs = 0.1:0.001:10;
fv = rand(size(xs));
ffun = @(x) interp1(xs, fv, x);
ic = [1; 2];
[t, y] = ode45(@(x, y) odeFun(x, y, ffun(x)), xspan, ic);
plot(t, y);

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Ameer Hamza on 17 Jun 2020
ic = [1; 2]
are initial condition and I choose them randomly. Note that I didn't start the solution at x=0, because, at x=0, there are singularities in your system. I started it at x=0.1. So the initial condition means
y(0.1) = 1;
y'(0.1) = 2;
You can choose these values according to the initial conditions of your system.
For the plotted lines, blue one corresponds to y(x), and the orange one is y'(x). the x-axis is x values of your equation. You can just plot y(x) using following line
plot(t, y(:,1));
PatrizioGraziosi on 17 Jun 2020
Brilliant! This is what I couldn't understand, that ic(2) is y', sorry, and many thanks for your support!
Ameer Hamza on 17 Jun 2020
I am glad to be of help!

David Goodmanson on 17 Jun 2020
Edited: David Goodmanson on 17 Jun 2020
Hi Patrezio,
d(log(x)) = dx/x, and you can insert that result in three locations to obtain
eq1 = x*diff( log(y), x) + x*diff( log(diff(y,x)), x) + x*diff( y, x)*log(x) + y == 1+f;
z = dsolve(eq1)
Warning: Unable to find explicit solution. Returning implicit solution instead.
> In dsolve (line 197)
solve([((C2 + f*y^2 + 2*y^2 - y^3)/(2*C1))^(1/(f - y + 2)) - x == 0, 1 < y - f], y) union ...
solve([((C2 + f*y^2 + 2*y^2 - y^3)/(2*C1))^(1/(f - y + 2)) - x == 0, ~1 < y - f], y)
There is no explicit solution for y(x), but there is a solution for x as a function of y. The solution is a union of two complementary regions of y, but if you are finding x as a function of y, that fact appears not to matter.

#### 1 Comment

PatrizioGraziosi on 17 Jun 2020
Hi David,
thank you!
Sure if I find a x as a function of y is fine, however, I find
syms y(x) f
eq = diff(log(y),x)*x + diff(log(diff(y,x)),x)*x + ...
diff(y,x)*x*log(x) + y ...
== 1 + f;
sol = dsolve(eq)
sol =
((2*C2*x^y + C2*f*x^y - C2*x^y*y + 2*C1*x^f*x^2)/(x^y*(f - y + 2)))^(1/2)
-((2*C2*x^y + C2*f*x^y - C2*x^y*y + 2*C1*x^f*x^2)/(x^y*(f - y + 2)))^(1/2)
which is specified in a different way from yours...