What is the solution for y^'-(2x+1)y+3x(5x-2)=0 ?

The solution for y^'-(2x+1)y+3x(5x-2)=0 is y=(integral of-3e^{-x^2-x}x(5x-2))/(e^{-x^2-x)}

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y^'-(2x+1)y+3x(5x-2)=0

y^{^{'}} - (2 x + 1) y + 3 x (5 x - 2) = 0

y = \frac{\int - 3 e ^{- x^{2} - x} x ( 5 x - 2 ) d x}{e ^{- x^{2} - x}}

Solution steps

y^{'} (x) - (2 x + 1) y + 3 x (5 x - 2) = 0

Solve linear ODE:

y = \frac{\int - 3 e ^{- x^{2} - x} x ( 5 x - 2 ) d x}{e ^{- x^{2} - x}}

y = \frac{\int - 3 e ^{- x^{2} - x} x ( 5 x - 2 ) d x}{e ^{- x^{2} - x}}

\frac{5}{x}, 1

\int_{2}^{7} \frac{1}{4 x + 8} d x

\int \frac{x + 1}{( x + 2 ) ( x - 3 )} d x

\int (tan (2 x) + cot (2 x))^{2} d x

\frac{6 x}{( 3 - 2 x ) ^{3}}

What is the solution for y^'-(2x+1)y+3x(5x-2)=0 ?
The solution for y^'-(2x+1)y+3x(5x-2)=0 is y=(integral of-3e^{-x^2-x}x(5x-2))/(e^{-x^2-x)}