function handle how to apply to dsolve???

syms t y
yp=@(t,y) y-t.^2+1;
exact=dsolve( 'Dy=yp(t,y)', 'y(0)=0.5');
abso=abs(diff(exact,2));

Explicit solution could not be found. 
> In dsolve (line 201)

how to change code????

Answer 1

[Walter Roberson]

syms y(t)
yp = y-t^2+1;
eqn = diff(y) == yp;
ic = y(0) == 1/2;
exact = dsolve([eqn, ic]);
abso = abs(diff(exact,2));

You are causing problems for yourself here and in the previous question by using y as both a function and a variable.

[Seong Ik Kim]

I want to remain function handle.. In your code there is no function handle, isnt it??

[Walter Roberson]

syms y t
yp = @(t,y) y-t.^2+1;
yt = sym('y(t)')
exact = dsolve( [diff(yt,t) == subs(yp(t,y), y, yt), 'y(0)=0.5'] );
abso = abs(diff(exact,2));

You are causing problems for yourself here and in the previous question by using y as both a function and a variable.

[Seong Ik Kim]

umm... this code has an error in 2017a. Support of character vectors that are not valid variable names or define a number will be removed in a future release. To create symbolic expressions, first create symbolic variables and then use operations on them.

[Walter Roberson]

That is not an error, that is a warning. Other than disabling the warning, there is no solution as long as you insist on using y as both a function and a variable name.

[Seong Ik Kim]

aha.. Thank you! and then, I have a quenstion. In symbolic , function handle is not recommend than symfun??

[Walter Roberson]

You have to invoke the function handle on appropriate symbolic variables in order to use it in dsolve() .

Your problem was a conflict between using y as a variable and y as a function.

You had

syms t y

which makes y a variable rather than a function.

You have

yp=@(t,y) y-t.^2+1;

when called with yp(t, y) this gives a result back in terms of the variable y. But in dsolve, you need the function y(t)

Ummm... I must be tired...

syms y(t)
yp = @(t,y) y-t.^2+1;
exact = dsolve([diff(y) == yp(t,y), y(0)==0.5]);
abso = abs(diff(exact,2));

[Seong Ik Kim]

Sorry, and Thank you. I understand my problems and your teaching. Thank you so much!!

Answer 2

[Torsten]

Use function handles if you want to integrate numerically, use symbolic expressions if you want to integrate symbolically.

Best wishes

Torsten.