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

The solution for y^'=(y^2+yx+x^2)/(x^2),y(1)=-1 is y=tan(ln(x)+(3pi)/4+pin)x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(y^2+yx+x^2)/(x^2),y(1)=-1

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

y = tan (ln (x) + \frac{3 π}{4} + πn) x

Solution steps

y^{'} (x) = \frac{y ^{2} + y x + x ^{2}}{x ^{2}}

Solve using substitution:

y = tan (ln (x) + \frac{3 π}{4} + πn) x

y = tan (ln (x) + \frac{3 π}{4} + πn) x

\frac{d}{d x} (e^{- 3 x} sin (2 x))

1 0^{8 - x^{2}}

\frac{\partial}{\partial x} ((x) (2 x y))

\frac{d y}{d x} = \frac{7 x}{4 y}

\int \frac{3}{cos ^{2} ( x )} d x

What is the solution for y^'=(y^2+yx+x^2)/(x^2),y(1)=-1 ?
The solution for y^'=(y^2+yx+x^2)/(x^2),y(1)=-1 is y=tan(ln(x)+(3pi)/4+pin)x