What is the solution for dy=(y^2-4)dx ?

The solution for dy=(y^2-4)dx is y=(2e^{4c_{1}})/(e^{4x)-e^{4c_{1}}}+(2e^{4x})/(e^{4x)-e^{4c_{1}}}

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dy=(y^2-4)dx

d y = (y^{2} - 4) d x

y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

Solution steps

d y = (y^{2} - 4) d x

Let

x

be the dependent variable. Divide by

d y

1 = (y^{2} - 4) \frac{d x}{d y}

Substitute

\frac{d x}{d y}

with

x^{^{'}}

1 = (y^{2} - 4) x^{^{^{'}}}

Isolate

x^{^{'}} : x^{^{'}} = \frac{1}{y ^{2} - 4}

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

then

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

x = \int \frac{1}{y ^{2} - 4} d y

\int \frac{1}{y ^{2} - 4} d y = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

x = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

Isolate

y : y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

h^{^{'}} = \frac{2 h}{x ln ( x )}

(y^{2} - 1) \frac{d x}{d y} = x

\frac{d y}{d x} + 12 x^{2} y = 5 x^{2}, y (0) = 4

y^{^{'}} - 2 x y - x = 0

x \frac{d y}{d x} = e^{4 y + x}

What is the solution for dy=(y^2-4)dx ?
The solution for dy=(y^2-4)dx is y=(2e^{4c_{1}})/(e^{4x)-e^{4c_{1}}}+(2e^{4x})/(e^{4x)-e^{4c_{1}}}