# Solve differential equation y'+y=x, \ y(0)=1

Question
Solve differential equation $$\displaystyle{y}'+{y}={x},\ {y}{\left({0}\right)}={1}$$

2021-03-09
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{d}{n}}}}+{y}={x}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{d}{n}}}}+{p}{y}={x}$$
$$\displaystyle{I}.{F}.={e}^{{\int{p}{\left.{d}{x}\right.}}}={e}^{{\int{1}{\left.{d}{x}\right.}}}={e}^{{x}}$$
$$\displaystyle{y}{\left({I}.{F}.\right)}=\int{0}{\left.{d}{x}\right.}+{c}$$
$$\displaystyle{e}^{{n}}{y}={\frac{{{x}^{{2}}}}{{{2}}}}+{c}$$
c=1
$$\displaystyle{y}={\frac{{{x}^{{2}}}}{{{2}}}}{e}^{{-{n}}}+{e}^{{-{x}}}$$

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