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Popular >

(dy)/(dx)=3y+y^2-2

Solution

\frac{d y}{d x} = 3 y + y^{2} - 2

Solution

y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

Solution steps

\frac{d y}{d x} = 3 y + y^{2} - 2

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = 3 y + y^{2} - 2

Solve for

x, y^{'} (x) = \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{x ^{^{'}}}

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

Isolate

x^{^{'}} : x^{^{'}} = \frac{1}{3 y + y ^{2} - 2}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x = \int \frac{1}{3 y + y ^{2} - 2} d y

\int \frac{1}{3 y + y ^{2} - 2} d y = - \frac{1}{17} (ln (\frac{2 y + 3}{17} + 1) - ln (\frac{2 y + 3}{17} - 1)) + c_{1}

x = - \frac{1}{17} (ln (\frac{2 y + 3}{17} + 1) - ln (\frac{2 y + 3}{17} - 1)) + c_{1}

Isolate

y : y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

Popular Examples

y^{^{'}} + (x + 8) y = 0

sec^2(x)tan(y)dx+sec^2(y)tan(x)dy=0

sec^{2} (x) tan (y) d x + sec^{2} (y) tan (x) d y = 0

\frac{y}{x} y^{^{'}} + e^{y} = 0

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(t^{2} - 22 t + 96) \frac{d y}{d t} = y

(dy}{dx}=\frac{xsin(x))/y

\frac{d y}{d x} = \frac{x sin ( x )}{y}

Frequently Asked Questions (FAQ)

What is the solution for (dy)/(dx)=3y+y^2-2 ?
The solution for (dy)/(dx)=3y+y^2-2 is