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If we consider the BVP $y^{\prime\prime}(t)=\frac{3}{2}y(t)^{2}$ where $0\leq t\leq1$ and $y(0)=4$ and $y(1)=1$. Then its solution is $y(t)=\frac{4}{(1+t)^{2}}$. I have set the BVP above as in mathemtica,
DSolve[{y''[t]==(3/2)(y[t])^2},y[0]==4,y[1]=1},y[x],x,0<= t <= 1]
When I run Mathemtica it says that DSolve::dsvar: $0\leq t\leq 1$ cannot be used as a variable.
$\begingroup$There are 6 typos in one line of your code. Using sol = DSolve[{y''[t] == (3/2) (y[t])^2}, y[t], t], we have out {{y[t] -> 2^(2/3) WeierstrassP[(t + C[1])/2^(2/3), {0, C[2]}]}}.$\endgroup$
$\begingroup$Dears, thanks for you help and suggesions. Please I have still issue in this equation solving. This example has been published in the paperhttps: //doi.org/10.1016/j.aml.2018.02.016$\endgroup$
$\begingroup$Here you can find solution to a more general equation ( setting $a=0,\; b=0,\; c=3/2$ you obtain your equation.) Symolically with DSolve one cannot solve BVP for this typ of ODE, however with a simple reasoning you can find the only one solution to the given problem. Nevertheless you incorrectly put 0 <= t <=1, one can evaluate e.g. DSolve[{y''[t] == (3/2) y[t]^2}, y[t], {t, 0, 1}].$\endgroup$
$\begingroup$Does this solve your problem How to solve a nonlinear second order ODE. In fact this is a duplicate for the reason mentioned above in my comment.$\endgroup$
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sol = DSolve[{y''[t] == (3/2) (y[t])^2}, y[t], t]
, we have out{{y[t] -> 2^(2/3) WeierstrassP[(t + C[1])/2^(2/3), {0, C[2]}]}}
. $\endgroup$DSolve
one cannot solve BVP for this typ of ODE, however with a simple reasoning you can find the only one solution to the given problem. Nevertheless you incorrectly put0 <= t <=1
, one can evaluate e.g.DSolve[{y''[t] == (3/2) y[t]^2}, y[t], {t, 0, 1}]
. $\endgroup$