What is the solution for x^7y^'=-8x^6y+cos(2x) ?

The solution for x^7y^'=-8x^6y+cos(2x) is y=(2xsin(2x)+cos(2x))/(4x^8)+(c_{1})/(x^8)

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x^7y^'=-8x^6y+cos(2x)

x^{7} y^{^{'}} = - 8 x^{6} y + cos (2 x)

y = \frac{2 x sin ( 2 x ) + cos ( 2 x )}{4 x ^{8}} + \frac{c _{1}}{x ^{8}}

Solution steps

x^{7} y^{'} (x) = - 8 x^{6} y + cos (2 x)

Solve linear ODE:

y = \frac{2 x sin ( 2 x ) + cos ( 2 x )}{4 x ^{8}} + \frac{c _{1}}{x ^{8}}

y = \frac{2 x sin ( 2 x ) + cos ( 2 x )}{4 x ^{8}} + \frac{c _{1}}{x ^{8}}

\int \frac{- 2 a x}{x ^{2} + y ^{2}} d y

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

\int x^{2} - 4 x d x

\int_{1}^{3} 3 r^{3} ln (r) d r

\int (7 x - 11)^{4} d x

What is the solution for x^7y^'=-8x^6y+cos(2x) ?
The solution for x^7y^'=-8x^6y+cos(2x) is y=(2xsin(2x)+cos(2x))/(4x^8)+(c_{1})/(x^8)