What is the solution for y^'=sqrt(-4y+57) ?

The solution for y^'=sqrt(-4y+57) is y=-t^2-2c_{1}t-c_{1}^2+57/4

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y^'=sqrt(-4y+57)

y^{^{'}} = - 4 y + 57

y = - t^{2} - 2 c_{1} t - c_{1}^{2} + \frac{57}{4}

Solution steps

y^{'} (t) = - 4 y + 57

Solve separable ODE:

y = - t^{2} - 2 c_{1} t - c_{1}^{2} + \frac{57}{4}

y = - t^{2} - 2 c_{1} t - c_{1}^{2} + \frac{57}{4}

e^{x} d x - y d y = 0, y (0) = 1

\int_{1}^{3} \frac{4}{x ^{2} + 1} d x

\int_{- 2}^{2} 4 (4 - x^{2}) d x

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x ^{2} - 9})

\int sec^{2} (\frac{7}{3} x + 4) d x

What is the solution for y^'=sqrt(-4y+57) ?
The solution for y^'=sqrt(-4y+57) is y=-t^2-2c_{1}t-c_{1}^2+57/4