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Fantom [35]
4 years ago
8

c{1}{6} x(1- y^{2} )" alt="y'= \frac{1}{6} x(1- y^{2} )" align="absmiddle" class="latex-formula">
This equation has the 2 constant solutions (in increasing order), y=? and y=?
The solution of this equation subject to the initial condition y(0)=7y(0)=7 is ?
Mathematics
1 answer:
svetlana [45]4 years ago
8 0
\dfrac{\mathrm dy}{\mathrm dx}=\dfrac16x(1-y^2)
\displaystyle\int\frac{\mathrm dy}{1-y^2}=\frac16\int x\,\mathrm dx
\dfrac12\ln|1+y|-\dfrac12\ln|1-y|=\dfrac1{12}x^2+C
\ln\left|\dfrac{1+y}{1-y}\right|=\dfrac16x^2+C
\dfrac{1+y}{1-y}=e^{x^2/6+C}
-1+\dfrac2{1-y}=Ce^{x^2/6}
y=1-\dfrac2{1+Ce^{x^2/6}}

Given that y(0)=7, we get

7=1-\dfrac2{1+Ce^0}\implies C=-\dfrac43

so the particular solution is

y=1-\dfrac2{1-\frac43e^{x^2/6}}
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