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How to solve in Mathematica this partial differential equation: $0.5\frac{\partial t(x,y)}{\partial x}+1.5\frac{\partial t(x,y)}{\partial y}+t(x,y)=y\cdot \sqrt{1+x^{3}}$ with condition $t(1,y)=y+2$?

I tried this:

DSolve[{0.5*D[t[x,y], x] + 1.5*D[t[x,y], y] + t[x,y] == y*Sqrt[1 + x^3]}, 
  t[1, y] == y + 2, t[x,y], {x,y}]

but after compilation I saw this message

DSolve::dsvar: "{a,b} cannot be used as a variable."

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    $\begingroup$ DSolve[{0.5*D[t[x, y], x] + 1.5*D[t[x, y], y] + t[x, y] == y*Sqrt[1 + x^3], t[1, y] == y + 2}, t[x, y], {x, y}]? $\endgroup$
    – Karsten7
    Dec 19, 2015 at 18:51
  • $\begingroup$ There is no a or b in your code, so presumably you have some leftover variable definitions. Try clearing them. $\endgroup$
    – user484
    Dec 19, 2015 at 18:59
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    $\begingroup$ You need to include the boundary condition t[1, y] == y + 2 inside the list of equations, not give it as a second argument. (This is what @Karsten7 showed but I thought it worth pointing out the change explicitly) $\endgroup$ Dec 19, 2015 at 20:24
  • $\begingroup$ Still I didn't receive solution. Now in output I've got the same thing as in input DSolve[{0.5*D[t[x, y], x] + 1.5*D[t[x, y], y] + t[x, y] == y*Sqrt[1 + x^3], t[1, y] == y + 2}, t[x, y], {x, y}] $\endgroup$
    – stitch
    Dec 20, 2015 at 8:46
  • $\begingroup$ @stitch. That indicates that Mathematica can't solve the differential equation analytically in the form that you've written it (and it's possible that there is no analytic solution). This means that you'll likely have to solve it numerically by using NDSolve. $\endgroup$
    – march
    Dec 20, 2015 at 18:04

1 Answer 1

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In 13.2 on Windows 10

DSolve[{1/2*D[t[x, y], x] + 3/2*D[t[x, y], y] + t[x, y] == 
y*Sqrt[1 + x^3], t[1, y] == y + 2}, t[x, y], {x, y}]

{{t[x, y] -> -E^(-2 x) (-5 E^2 + 3 E^2 x - E^2 y + Inactive[Integrate][ 2 E^(2 K[1]) (-3 x + y + 3 K[1]) Sqrt[1 + K[1]^3], {K[1], 1, 1}] - Inactive[Integrate][ 2 E^(2 K[1]) (-3 x + y + 3 K[1]) Sqrt[1 + K[1]^3], {K[1], 1, x}])}}

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