What is the solution for xy^'-y=2020x^2,y(1)=1 ?

The solution for xy^'-y=2020x^2,y(1)=1 is y=2020x^2-2019x

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xy^'-y=2020x^2,y(1)=1

x y^{^{'}} - y = 2020 x^{2}, y (1) = 1

y = 2020 x^{2} - 2019 x

Solution steps

x y^{'} (x) - y = 2020 x^{2}

Solve linear ODE:

y = 2020 x^{2} - 2019 x

y = 2020 x^{2} - 2019 x

\int 12 sin^{3} (x co) s^{2} x d x

y^{^{'}} - \frac{1}{x} y = 9, y (1) = \frac{1}{7}

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

\frac{d}{d x} (x^{3} - 5 x)

\frac{d}{d x} ((1 - x) (x^{2} - 2)^{2})

What is the solution for xy^'-y=2020x^2,y(1)=1 ?
The solution for xy^'-y=2020x^2,y(1)=1 is y=2020x^2-2019x