What is the solution for x^'=x^2-4 ?

The solution for x^'=x^2-4 is x=(2e^{4c_{1}})/(e^{4t)-e^{4c_{1}}}+(2e^{4t})/(e^{4t)-e^{4c_{1}}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^'=x^2-4

x^{^{'}} = x^{2} - 4

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

Solution steps

x^{'} (t) = x^{2} - 4

Solve for

t, x^{'} (t) = \frac{d x}{d t} = \frac{1}{\frac{d t}{d x}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = x^{2} - 4

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{x ^{2} - 4}

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

then

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

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

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

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

Isolate

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

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

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