What is the solution for sqrt(1-4x^2)y^'=x ?

The solution for sqrt(1-4x^2)y^'=x is y=-1/4 sqrt(1-4x^2)+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(1-4x^2)y^'=x

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

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

Solution steps

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

Isolate

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

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

then

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

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

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

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

\frac{x}{2}

x \to 0 lim (x^{11 x})

\frac{d P}{d t} = (\frac{9}{240}) P - \frac{1}{4800} (P)^{2}

x \frac{d y}{d x} + y - y^{2} e^{2 x} = 0

x \to 0 lim (sin (x) \cdot ln (x))

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