y^'=(18(1+3y))/((3-3x)^2),y(2)=0

Lösung

y^{^{'}} = \frac{18 ( 1 + 3 y )}{( 3 - 3 x ) ^{2}}, y (2) = 0

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

Schritte zur Lösung

y^{'} (x) = \frac{18 ( 1 + 3 y )}{( 3 - 3 x ) ^{2}}

Solve separable ODE:

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

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