What is the solution for 25yy^'-4x=0 ?

The solution for 25yy^'-4x=0 is y=(sqrt(4x^2+c_{1)})/5 ,y=-(sqrt(4x^2+c_{1)})/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25yy^'-4x=0

25 y y^{^{'}} - 4 x = 0

y = \frac{4 x ^{2} + c _{1}}{5}, y = - \frac{4 x ^{2} + c _{1}}{5}

Solution steps

25 y y^{'} (x) - 4 x = 0

Solve separable ODE:

y = \frac{4 x ^{2} + c _{1}}{5}, y = - \frac{4 x ^{2} + c _{1}}{5}

y = \frac{4 x ^{2} + c _{1}}{5}, y = - \frac{4 x ^{2} + c _{1}}{5}

\frac{d}{d x} (ln (- 1))

\int (x^{2} - 4)^{3} 2 x d x

\frac{- 12}{( 1 - x ) ^{4}}

\int sinh (8 x) d x

\frac{d}{d x} (x x^{2} + 36)

What is the solution for 25yy^'-4x=0 ?
The solution for 25yy^'-4x=0 is y=(sqrt(4x^2+c_{1)})/5 ,y=-(sqrt(4x^2+c_{1)})/5