What is the solution for x^2y^'-4xy=sin(3x)x^6 ?

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

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x^2y^'-4xy=sin(3x)x^6

x^{2} y^{^{'}} - 4 x y = sin (3 x) x^{6}

y = - \frac{1}{3} cos (3 x) x^{4} + c_{1} x^{4}

Solution steps

x^{2} y^{'} (x) - 4 x y = sin (3 x) x^{6}

Solve linear ODE:

y = - \frac{1}{3} cos (3 x) x^{4} + c_{1} x^{4}

y = - \frac{1}{3} cos (3 x) x^{4} + c_{1} x^{4}

f (x) = 2 sec (x) tan (x)

\int \frac{( 6 + e ^{x} ) ^{2}}{e ^{x}} d x

f (x) = e^{x} + ln (x + 1)

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

k = 1 \sum \infty k^{- 2.4}

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