What is the solution for (1-x^3)y^'-2(1+x)y=y^{5/2} ?

The solution for (1-x^3)y^'-2(1+x)y=y^{5/2} is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1-x^3)y^'-2(1+x)y=y^{5/2}

(1 - x^{3}) y^{^{'}} - 2 (1 + x) y = y^{\frac{5}{2}}

y = \frac{2 3 2 3 x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1}{3 9 - 24 c _{1} x ^{2} + 48 c _{1} x - 24 c _{1} + 16 c _{1}^{2} x ^{4} - 64 c _{1}^{2} x ^{3} + 96 c _{1}^{2} x ^{2} - 64 c _{1}^{2} x + 16 c _{1}^{2}}

Solution steps

(1 - x^{3}) y^{'} (x) - 2 (1 + x) y = y^{\frac{5}{2}}

Solve Bernoulli ODE:

y = \frac{2 3 2 3 x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1}{3 9 - 24 c _{1} x ^{2} + 48 c _{1} x - 24 c _{1} + 16 c _{1}^{2} x ^{4} - 64 c _{1}^{2} x ^{3} + 96 c _{1}^{2} x ^{2} - 64 c _{1}^{2} x + 16 c _{1}^{2}}

y = \frac{2 3 2 3 x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1}{3 9 - 24 c _{1} x ^{2} + 48 c _{1} x - 24 c _{1} + 16 c _{1}^{2} x ^{4} - 64 c _{1}^{2} x ^{3} + 96 c _{1}^{2} x ^{2} - 64 c _{1}^{2} x + 16 c _{1}^{2}}

\int 8 + 6 x + 24 x^{2} d x

\frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}})

\frac{\partial}{\partial x} (\frac{4 x + 9}{3 y + 1})

f (x) = 4 x^{2} + 5 x + 3

\int (x^{3} + 3 x)^{32} (3 x^{2} + 3) d x

What is the solution for (1-x^3)y^'-2(1+x)y=y^{5/2} ?
The solution for (1-x^3)y^'-2(1+x)y=y^{5/2} is