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I want to prove that the followning Cauchy problem doesn't have solution:

$y'(x)\sin(x)=y\ln(y),y(\frac{\pi}{2})=1$

I solved the ODE and i found:

$\ln(|\ln(y)|)=\ln(|\tan(\frac{x}{2})|)+c$

When i put the initial condition, i found ln(0). How do i prove, that this problem doesn't have solution?

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    $\begingroup$ This site is about the software Mathematica, your question seem to be about the mathematics, You should ask your question here: math.stackexchange.com probably... $\endgroup$
    – mattiav27
    Oct 14 at 13:38
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The general solution is

Y = DSolveValue[{y'[x] Sin[x] == y[x] Log[y[x]]  }, y, x,GeneratedParameters -> k]    
(*Function[{x}, E^(E^k[1] Tan[x/2])]*)

Plot shows

Plot[ {Y[Pi/2 ], 1}, {k[1], -1, 1}, PlotRange -> {0, Automatic}]

enter image description here

that Y[Pi/2]>1! The boundary condition Y[Pi/2]==1 can't be satisfied, the solution doesn't exist in this case!

The mathematical prove follows via Limit[Y[Pi/2],k[1]->-Infinity] (*1*) and Y'[x]>0]

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