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Question Number 543 by 123456 last updated on 25/Jan/15
if f′(x)=f(x)+x, f(0)=0, then f(1)=?
iff′(x)=f(x)+x,f(0)=0,thenf(1)=?
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
y=f(x)y′−y=xI.F.=e∫−1dx=e−xe−xy′−e−xy=xe−xye−x=∫xe−xdxye−x=−xe−x−e−x+Cy=−x−1+Cexx=0y(0)=−1+C⇒C=1y=−(x+1)+exy(1)=−2+e

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