What is the solution for dx=(e^{t^2}+2xt)dt ?

The solution for dx=(e^{t^2}+2xt)dt is x=e^{t^2}t+c_{1}e^{t^2}

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dx=(e^{t^2}+2xt)dt

d x = (e^{t^{2}} + 2 x t) d t

x = e^{t^{2}} t + c_{1} e^{t^{2}}

Solution steps

d x = (e^{t^{2}} + 2 x t) d t

Solve linear ODE:

x = e^{t^{2}} t + c_{1} e^{t^{2}}

x = e^{t^{2}} t + c_{1} e^{t^{2}}

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\frac{d}{d x} (9 cos (x) - 7 x^{2})

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What is the solution for dx=(e^{t^2}+2xt)dt ?
The solution for dx=(e^{t^2}+2xt)dt is x=e^{t^2}t+c_{1}e^{t^2}