What is the solution for y^{''}=(2k)/(1-k) ?

The solution for y^{''}=(2k)/(1-k) is y=(kt^2)/(1-k)+c_{1}t+c_{2}

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y^{''}=(2k)/(1-k)

y^{^{''}} = \frac{2 k}{1 - k}

y = \frac{k t ^{2}}{1 - k} + c_{1} t + c_{2}

Solution steps

y^{''} (t) = \frac{2 k}{1 - k}

Solve

y^{''} (t) = \frac{2 k}{1 - k} : y^{'} (t) = \frac{2 k t}{1 - k} + c_{1}

Solve

y^{'} (t) = \frac{2 k t}{1 - k} + c_{1} : y = \frac{k t ^{2}}{1 - k} + c_{1} t + c_{2}

y = \frac{k t ^{2}}{1 - k} + c_{1} t + c_{2}

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f (x) = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}

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x \to 7 lim (\frac{- 2}{7 - x})

y = \frac{6 x}{x + 2}, (4, 4)

What is the solution for y^{''}=(2k)/(1-k) ?
The solution for y^{''}=(2k)/(1-k) is y=(kt^2)/(1-k)+c_{1}t+c_{2}