What is the solution for y^'=(4x^3+1)/(2y-6),y(1)=2 ?

The solution for y^'=(4x^3+1)/(2y-6),y(1)=2 is y=3-sqrt(x^4+x-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(4x^3+1)/(2y-6),y(1)=2

y^{^{'}} = \frac{4 x ^{3} + 1}{2 y - 6}, y (1) = 2

y = 3 - x^{4} + x - 1

Solution steps

y^{'} (x) = \frac{4 x ^{3} + 1}{2 y - 6}

Solve separable ODE:

y = 3 - x^{4} + x - 1

y = 3 - x^{4} + x - 1

\frac{d}{d x} (a^{2} x^{2})

\int 3 cos (5 x) d x

x \to \infty lim (an)

6 4 x - \frac{5}{x ^{2}}

\int \frac{x + 36}{x} d x

What is the solution for y^'=(4x^3+1)/(2y-6),y(1)=2 ?
The solution for y^'=(4x^3+1)/(2y-6),y(1)=2 is y=3-sqrt(x^4+x-1)