What is the solution for (x-3)(dy)/(dx)=4(x-3)^4+y ?

The solution for (x-3)(dy)/(dx)=4(x-3)^4+y is y=(4x^4)/3-16x^3+72x^2-108x+c_{1}x-3c_{1}

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(x-3)(dy)/(dx)=4(x-3)^4+y

(x - 3) \frac{d y}{d x} = 4 (x - 3)^{4} + y

y = \frac{4 x ^{4}}{3} - 16 x^{3} + 72 x^{2} - 108 x + c_{1} x - 3 c_{1}

Solution steps

(x - 3) \frac{d y}{d x} = 4 (x - 3)^{4} + y

Solve linear ODE:

y = \frac{4 x ^{4}}{3} - 16 x^{3} + 72 x^{2} - 108 x + c_{1} x - 3 c_{1}

y = \frac{4 x ^{4}}{3} - 16 x^{3} + 72 x^{2} - 108 x + c_{1} x - 3 c_{1}

\frac{\partial}{\partial y} (e^{y x^{2}})

\int \frac{70}{( x + 1 ) ( x ^{2} + 9 ) ^{2}} d x

x \to 1 + lim (\frac{x + 7}{x + 2})

x - ln (x)

\frac{d y}{d x}, y = 2 x

What is the solution for (x-3)(dy)/(dx)=4(x-3)^4+y ?
The solution for (x-3)(dy)/(dx)=4(x-3)^4+y is y=(4x^4)/3-16x^3+72x^2-108x+c_{1}x-3c_{1}