# Did you know?: x% of y = y% of x

The exercise is about lớn solve sầu \$y""=1/y^3\$ for \$y=y(t)\$ with initial values \$y(0)=1\$ và \$y"(0)=2\$.

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My attempts:

I wrote up it as a first order diff.equation with \$2\$ variables by introducing \$x:=y"\$, so that we are looking for (the \$y\$-coordinate of) the trajectory of the vector field \$(1/y^3, x)\$, I could draw some vectors, and could "guess" how the solution will look lượt thích.

Look for the solution in size \$y=c,t^alpha\$. I got that \$alpha=1/2\$ and \$c^4=-4\$, that is \$c=pm1pm i\$. However, I guess, real function was asked.

I guess I"m missing something I should know about solving such a differential equation.

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ordinary-differential-equations
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asked May 31 "13 at 7:41

z17z17
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Multiply \$y"\$,\$\$y"y""=y^-3y"\$\$Integrate, \$\$frac 1 2 (y")^2 = -frac 1 2 y^-2 + c\$\$Using initial condition, \$\$2=-frac 1 2 +c\$\$Thus \$c=frac 5 2\$. Thus, we have \$\$(y")^2+y^-2=5.\$\$Multiply \$y^2\$, and substitute \$Y=y^2\$, then\$\$frac (Y")^2 4 +1 = 5Y,\$\$\$\$(Y")^2=20Y-4\$\$I think it is easy lớn go from here.

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answered May 31 "13 at 7:59
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Sungjin Kyên ổn
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