What is the solution for y^'=(3x-2)^2(5x+4) ?

The solution for y^'=(3x-2)^2(5x+4) is y=(45x^4)/4-8x^3-14x^2+16x+c_{1}

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y^'=(3x-2)^2(5x+4)

y^{^{'}} = (3 x - 2)^{2} (5 x + 4)

y = \frac{45 x ^{4}}{4} - 8 x^{3} - 14 x^{2} + 16 x + c_{1}

Solution steps

y^{'} (x) = (3 x - 2)^{2} (5 x + 4)

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

then

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

y = \int (3 x - 2)^{2} (5 x + 4) d x

\int (3 x - 2)^{2} (5 x + 4) d x = \frac{45 x ^{4}}{4} - 8 x^{3} - 14 x^{2} + 16 x + c_{1}

y = \frac{45 x ^{4}}{4} - 8 x^{3} - 14 x^{2} + 16 x + c_{1}

t^{3} - 12 t

y^{^{''}} = \frac{1}{x ^{2}} y

\int_{0}^{x} sin (t) d t

\int_{1}^{3} (x^{2} - 1) d x

\int_{1}^{100} \frac{1}{1 + x ^{2}} d x

What is the solution for y^'=(3x-2)^2(5x+4) ?
The solution for y^'=(3x-2)^2(5x+4) is y=(45x^4)/4-8x^3-14x^2+16x+c_{1}