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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}]? $\endgroup$ – Karsten 7. Dec 19 '15 at 18:51
  • $\begingroup$ There is no a or b in your code, so presumably you have some leftover variable definitions. Try clearing them. $\endgroup$ – Rahul Dec 19 '15 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) $\endgroup$ – Simon Woods Dec 19 '15 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 '15 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 '15 at 18:04

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