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I wanted to solve a differential equation involving a term with integration over its parameters. I want to call the integral expression but I get the wrong result:

 ysol4 = ParametricNDSolveValue[{D[y[t], t] == y[t]*x + 
Integrate[y[t], {x, 0, 1}], 
y[0] == 1}, Integrate[y[t], {x, 0, 1}], {t, 0, 30}, {x}];
Plot[Evaluate[ysol4[0.1]], {t, 0, 1}]

I know that the result is wrong since the plot is sensitive to x. It should not be since I called the expression that is an integral of x. Any help will be appreciated.

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  • $\begingroup$ It's because Integrate[y[t], {x, 0, 1}] is just y[t] - in other words, it's treating y[t] like a constant in the integral. $\endgroup$
    – flinty
    Aug 24, 2020 at 16:21

2 Answers 2

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Generally speaking, we have here integrodifferential equation, so function $y=y(x,t)$, and we can consider equation and its solution as follows

ysol = NDSolveValue[{D[y[x, t], t] == 
    y[x, t]*x + Integrate[y[x, t], {x, 0, 1}], y[x, 0] == 1}, 
  y[x, t], {t, 0, 30}, {x, 0, 1}]

It looks like NDSolve can recognize and solve it (?), and we can visualize solution as

Plot3D[ysol, {t, 0, 1}, {x, .0, 1}, AxesLabel -> Automatic,  
 Mesh -> None, ColorFunction -> "Rainbow"]

Figure 1

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  • $\begingroup$ This seems to work. Thanks! $\endgroup$
    – khvillegas
    Aug 25, 2020 at 4:01
  • $\begingroup$ @khvillegas You are welcome! $\endgroup$
    – Alex Trounev
    Aug 25, 2020 at 10:15
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How about this:

DSolve[{D[y[t], t] == y[t]*x + 
Integrate[y[t], {x, 0, 1}], 
y[0] == 1},y[t],t]
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