What is the solution for 3xy^'-15y=x^{-4},y(1)=-2 ?

The solution for 3xy^'-15y=x^{-4},y(1)=-2 is y=-1/(27x^4)-(53x^5)/(27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3xy^'-15y=x^{-4},y(1)=-2

3 x y^{^{'}} - 15 y = x^{- 4}, y (1) = - 2

y = - \frac{1}{27 x ^{4}} - \frac{53 x ^{5}}{27}

Solution steps

3 x y^{'} (x) - 15 y = x^{- 4}

Solve linear ODE:

y = - \frac{1}{27 x ^{4}} - \frac{53 x ^{5}}{27}

y = - \frac{1}{27 x ^{4}} - \frac{53 x ^{5}}{27}

\int_{0}^{4.5} 90 x d x

\frac{d}{d x} (1 + 2 tan (x))

x \to 0 lim (x^{2} - 1)

\int ln (x) x^{15} d x

\frac{3}{( 4 x + 1 ) ^{2}}

What is the solution for 3xy^'-15y=x^{-4},y(1)=-2 ?
The solution for 3xy^'-15y=x^{-4},y(1)=-2 is y=-1/(27x^4)-(53x^5)/(27)