What is the solution for x(dy)/(dx)-4y=x^6x^x ?

The solution for x(dy)/(dx)-4y=x^6x^x is y=x^4*integral of x^{x+1}

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x(dy)/(dx)-4y=x^6x^x

x \frac{d y}{d x} - 4 y = x^{6} x^{x}

y = x^{4} \cdot \int x^{x + 1} d x

Solution steps

x \frac{d y}{d x} - 4 y = x^{6} x^{x}

Solve linear ODE:

y = x^{4} \cdot \int x^{x + 1} d x

y = x^{4} \cdot \int x^{x + 1} d x

\int 50 + 3.5 x e^{- 0.01 x^{2}} d x

y^{^{'}} = x e^{x^{2} - y}

y^{^{'}} = \frac{( 3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2} )}{x ^{5} y}

x y^{^{'}} = y + x^{2} e^{2 x}

What is the solution for x(dy)/(dx)-4y=x^6x^x ?
The solution for x(dy)/(dx)-4y=x^6x^x is y=x^4*integral of x^{x+1}