What is the solution for 4x^3y+x^4y^'=-6sec^2(x)tan(x) ?

The solution for 4x^3y+x^4y^'=-6sec^2(x)tan(x) is y=-(3tan^2(x))/(x^4)+(c_{1})/(x^4)

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4x^3y+x^4y^'=-6sec^2(x)tan(x)

4 x^{3} y + x^{4} y^{^{'}} = - 6 sec^{2} (x) tan (x)

y = - \frac{3 tan ^{2} ( x )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

4 x^{3} y + x^{4} y^{'} (x) = - 6 sec^{2} (x) tan (x)

Solve linear ODE:

y = - \frac{3 tan ^{2} ( x )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

y = - \frac{3 tan ^{2} ( x )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

\frac{\partial}{\partial x} (A e^{\frac{1}{x}})

\int_{0}^{6} 2 π (y + 2) (3 - \frac{y}{2}) d y

\frac{d y}{d x} - \frac{y}{x} = 3 x e^{x}

\frac{d}{d x} (ln (x e^{- 5 x}))

\int_{- 2}^{2} (\frac{1}{( x + 4 ) ( x - 4 )}) d x

What is the solution for 4x^3y+x^4y^'=-6sec^2(x)tan(x) ?
The solution for 4x^3y+x^4y^'=-6sec^2(x)tan(x) is y=-(3tan^2(x))/(x^4)+(c_{1})/(x^4)