What is the solution for 8x^7y+x^8y^'=-5sec^2(x)tan(x) ?

The solution for 8x^7y+x^8y^'=-5sec^2(x)tan(x) is y=-(5tan^2(x))/(2x^8)+(c_{1})/(x^8)

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8x^7y+x^8y^'=-5sec^2(x)tan(x)

8 x^{7} y + x^{8} y^{^{'}} = - 5 sec^{2} (x) tan (x)

y = - \frac{5 tan ^{2} ( x )}{2 x ^{8}} + \frac{c _{1}}{x ^{8}}

Solution steps

8 x^{7} y + x^{8} y^{'} (x) = - 5 sec^{2} (x) tan (x)

Solve linear ODE:

y = - \frac{5 tan ^{2} ( x )}{2 x ^{8}} + \frac{c _{1}}{x ^{8}}

y = - \frac{5 tan ^{2} ( x )}{2 x ^{8}} + \frac{c _{1}}{x ^{8}}

\frac{d}{d x} (- x + y + 2)

x \to \frac{π}{6} lim (\frac{sin ( x ) - sin ( \frac{π}{6} )}{x - \frac{π}{6}})

\int 20 tan (2 x) sec^{2} (2 x) d x

\frac{d}{d x} (sin (ln (4 x)))

e^{- 4 x^{2}}

What is the solution for 8x^7y+x^8y^'=-5sec^2(x)tan(x) ?
The solution for 8x^7y+x^8y^'=-5sec^2(x)tan(x) is y=-(5tan^2(x))/(2x^8)+(c_{1})/(x^8)