What is the solution for 2yy^'+4=y^2+4x,y(0)=5 ?

The solution for 2yy^'+4=y^2+4x,y(0)=5 is y=sqrt(-4x+25e^x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2yy^'+4=y^2+4x,y(0)=5

2 y y^{^{'}} + 4 = y^{2} + 4 x, y (0) = 5

y = - 4 x + 25 e^{x}

Solution steps

2 y y^{'} (x) + 4 = y^{2} + 4 x

Solve Bernoulli ODE:

y = - 4 x + 25 e^{x}

y = - 4 x + 25 e^{x}

t \to 0 lim (sin (3 t))

\int \frac{4 - x ^{2}}{2} d x

\frac{d}{d x} (x^{3} - 9 x^{2} + 24 x)

n = 1 \sum \infty \frac{2 ^{n}}{n ^{2}}

\frac{1}{2 x} - \frac{5}{3 x ^{2}}

What is the solution for 2yy^'+4=y^2+4x,y(0)=5 ?
The solution for 2yy^'+4=y^2+4x,y(0)=5 is y=sqrt(-4x+25e^x)