Question

: Find the general solution of y′−2xy=1

Answer 1

Answer on Question #69601 – Math – Differential Equations

Question

Find the general solution of $y' - 2xy = 1$

Solution

$$y' + P(x)y = Q(x)$$ $$y = e^{-\int P(x)dx} \left( \int Q(x) e^{\int P(x)dx} dx + C \right)$$ $$\frac{dy}{dx} - 2xy = 1, \quad P(x) = -2x, \quad Q(x) = 1$$ $$e^{-\int P(x)dx} = e^{-\int -2xdx} = e^{x^2}$$ $$e^{\int P(x)dx} = e^{\int -2xdx} = e^{-x^2}$$ $$y = e^{x^2} \cdot \left( \int e^{-x^2} dx + C \right)$$ $$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt$$ $$y = e^{x^2} \cdot \left( \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + C \right)$$

Answer:

$$y = e^{x^2} \cdot \left( \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + C \right).$$

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