What is the solution for y^'=0.4(10-x)(5-x) ?

The solution for y^'=0.4(10-x)(5-x) is y=0.4(50x-7.5x^2+0.33333…x^3)+c_{1}

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y^'=0.4(10-x)(5-x)

y^{^{'}} = 0.4 (10 - x) (5 - x)

y = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}

Solution steps

y^{'} (x) = 0.4 (10 - x) (5 - x)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 0.4 (10 - x) (5 - x) d x

\int 0.4 (10 - x) (5 - x) d x = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}

y = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}

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What is the solution for y^'=0.4(10-x)(5-x) ?
The solution for y^'=0.4(10-x)(5-x) is y=0.4(50x-7.5x^2+0.33333…x^3)+c_{1}