What is the solution for (((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y) ?

The solution for (((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y)

(\frac{( x + 1 ) ^{2} + 6}{4 ( x + 1 )}) (\frac{d y}{d x}) = tan (y)

y = arcsin (e^{c_{1}} x^{4} + 4 e^{c_{1}} x^{3} + 18 e^{c_{1}} x^{2} + 28 e^{c_{1}} x + 49 e^{c_{1}}) + 2 πn, y = π + arcsin (- e^{c_{1}} x^{4} - 4 e^{c_{1}} x^{3} - 18 e^{c_{1}} x^{2} - 28 e^{c_{1}} x - 49 e^{c_{1}}) + 2 πn

Solution steps

(\frac{( x + 1 ) ^{2} + 6}{4 ( x + 1 )}) (\frac{d y}{d x}) = tan (y)

Solve separable ODE:

y = arcsin (e^{c_{1}} x^{4} + 4 e^{c_{1}} x^{3} + 18 e^{c_{1}} x^{2} + 28 e^{c_{1}} x + 49 e^{c_{1}}) + 2 πn, y = π + arcsin (- e^{c_{1}} x^{4} - 4 e^{c_{1}} x^{3} - 18 e^{c_{1}} x^{2} - 28 e^{c_{1}} x - 49 e^{c_{1}}) + 2 πn

y = arcsin (e^{c_{1}} x^{4} + 4 e^{c_{1}} x^{3} + 18 e^{c_{1}} x^{2} + 28 e^{c_{1}} x + 49 e^{c_{1}}) + 2 πn, y = π + arcsin (- e^{c_{1}} x^{4} - 4 e^{c_{1}} x^{3} - 18 e^{c_{1}} x^{2} - 28 e^{c_{1}} x - 49 e^{c_{1}}) + 2 πn

\frac{d y}{d x} = - \frac{a y}{x}

y \cdot y^{^{'}} - 2 x = sin (x)

\frac{d y}{d x} = 0.125 (y - 3), y (0) = 5

e^{x} \frac{d y}{d x} - sec (y) - x sec (y) = 0

2 y y^{^{'}} + 1 = e^{x - 1}

What is the solution for (((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y) ?
The solution for (((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y) is