differential equation with mixed linear and log derivatives - proper setting

Question

PatrizioGraziosi 2020-6-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/549624-differential-equation-with-mixed-linear-and-log-derivatives-proper-setting

评论： Ameer Hamza 2020-6-17

采纳的回答： Ameer Hamza

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;

I always get an error about the log in the differentiation

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.

Can someone help me, please?

Thanks,

Patrizio

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Answer 1

Ameer Hamza 2020-6-17

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https://ww2.mathworks.cn/matlabcentral/answers/549624-differential-equation-with-mixed-linear-and-log-derivatives-proper-setting#answer_452409

编辑：Ameer Hamza 2020-6-17

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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PatrizioGraziosi 2020-6-17

Brilliant! This is what I couldn't understand, that ic(2) is y', sorry, and many thanks for your support!

Ameer Hamza 2020-6-17

I am glad to be of help!

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Answer 2

David Goodmanson 2020-6-17

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https://ww2.mathworks.cn/matlabcentral/answers/549624-differential-equation-with-mixed-linear-and-log-derivatives-proper-setting#answer_452316

编辑：David Goodmanson 2020-6-17

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.

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PatrizioGraziosi 2020-6-17

编辑：PatrizioGraziosi 2020-6-17

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)