Question Number 543 by 123456 last updated on 25/Jan/15
![if f′(x)=f(x)+x, f(0)=0, then f(1)=?](https://www.tinkutara.com/question/Q543.png)
$${if}\:{f}'\left({x}\right)={f}\left({x}\right)+{x},\:{f}\left(\mathrm{0}\right)=\mathrm{0},\:{then} \\ $$$${f}\left(\mathrm{1}\right)=? \\ $$
Answered by prakash jain last updated on 24/Jan/15
![y=f(x) y′−y=x I.F.=e^(∫−1dx) =e^(−x) e^(−x) y′−e^(−x) y=xe^(−x) ye^(−x) =∫xe^(−x) dx ye^(−x) =−xe^(−x) −e^(−x) +C y=−x−1+Ce^x x=0 y(0)=−1+C⇒C=1 y=−(x+1)+e^x y(1)=−2+e](https://www.tinkutara.com/question/Q546.png)
$${y}={f}\left({x}\right) \\ $$$${y}'−{y}={x} \\ $$$${I}.{F}.={e}^{\int−\mathrm{1}{dx}} ={e}^{−{x}} \\ $$$${e}^{−{x}} {y}'−{e}^{−{x}} {y}={xe}^{−{x}} \\ $$$${ye}^{−{x}} =\int{xe}^{−{x}} {dx} \\ $$$${ye}^{−{x}} =−{xe}^{−{x}} −{e}^{−{x}} +{C} \\ $$$${y}=−{x}−\mathrm{1}+{Ce}^{{x}} \\ $$$${x}=\mathrm{0} \\ $$$${y}\left(\mathrm{0}\right)=−\mathrm{1}+{C}\Rightarrow{C}=\mathrm{1} \\ $$$${y}=−\left({x}+\mathrm{1}\right)+{e}^{{x}} \\ $$$${y}\left(\mathrm{1}\right)=−\mathrm{2}+{e} \\ $$