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Question Number 140896 by bramlexs22 last updated on 14/May/21
thefunctionfwithvariablex satisfiestheequation x2f′(x)+2xf(x)=arctanxfor 0<arctanx<π2andf(1)=π4. findf(x).
Answered by EDWIN88 last updated on 15/May/21
letg(x)=x2f(x),diffwrtxgive g′(x)=x2f′(x)+2xf(x) sothatg′(x)=arctanx integrationbypart g(x)=∫arctanxdx=xarctanx−12ln(1+x2)+c theconditiong(1)=f(1)=π4 ⇒π4=arctan1−12ln(2)+c ⇒c=ln2 ∴f(x)=g(x)x2=arctanxx−ln1+x2x2+ln2x2
Answered by mathmax by abdo last updated on 14/May/21
f(x)=ye⇒x2y′+2xy=arctanxandy(1)=π4 h→xy′+2y=0⇒xy′=−2y⇒y′y=−2x⇒ln∣y∣=−2log∣x∣+c⇒ y=kx2mvcmethod→y′=k′x2+k(−2xx4)=k′x2−2kx3 e⇒k′−2kx+2kx=arctanx⇒k=∫arctanxdx =xarctanx−∫x1+x2dx=xarctan(x)−12log(1+x2)+C⇒ y(x)=1x2(xarctan(x)−12log(1+x2)+C) =arctan(x)x−12x2log(1+x2)+cx2=f(x) f(1)=π4⇒π4−12log2+C=π4⇒C=12log2⇒ f(x)=arctan(x)x−12x2log(1+x2)+log22x2
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