Menu Close

Given-f-x-0-1-x-y-2-f-y-dy-2x-2-3x-1-find-f-x-




Question Number 189066 by cortano12 last updated on 11/Mar/23
 Given f(x)+∫_0 ^1 (x+y)^2  f(y) dy=2x^2 −3x+1   find f(x).
Givenf(x)+10(x+y)2f(y)dy=2x23x+1findf(x).
Answered by horsebrand11 last updated on 11/Mar/23
2x^2 −3x+1=f(x)+∫_0 ^1 (x^2 +2xy+y^2 )dy  2x^2 −3x+1=f(x)+x^2 ∫_0 ^1 f(y)dy+2x∫_0 ^1 y f(y)dy+∫_0 ^1 y^2  f(y) dy  let  { ((p=∫_0 ^1 f(y)dy)),((q=∫_0 ^1 y f(y)dy )),((r=∫_0 ^1 y^2  f(y)dy )) :}  f(x)=2x^2 −px^2 −3x−2qx+1−r  f(x)=(2−p)x^2 −(3+2q)x+(1−r)
2x23x+1=f(x)+10(x2+2xy+y2)dy2x23x+1=f(x)+x210f(y)dy+2x10yf(y)dy+10y2f(y)dylet{p=10f(y)dyq=10yf(y)dyr=10y2f(y)dyf(x)=2x2px23x2qx+1rf(x)=(2p)x2(3+2q)x+(1r)
Answered by mr W last updated on 11/Mar/23
f(x)+∫_0 ^1 (x+y)^2 f(y)dy=2x^2 −3x+1  f(x)+∫_0 ^1 (x^2 +2xy+y^2 )f(y)dy=2x^2 −3x+1  f(x)+x^2 ∫_0 ^1 f(y)dy+2x∫_0 ^1 yf(y)dy+∫_0 ^1 y^2 f(y)dy=2x^2 −3x+1  f(x)+Ax^2 +2Bx+C=2x^2 −3x+1  ⇒f(x)=(2−A)x^2 −(3+2B)x+(1−C)  A=∫_0 ^1 f(y)dy=∫_0 ^1 [(2−A)y^2 −(3+2B)y+(1−C)]dy  A=(2−A)(1/3)−(3+2B)(1/2)+(1−C)  ⇒8A+6B+6C=1   ...(i)  B=∫_0 ^1 yf(y)dy=∫_0 ^1 [(2−A)y^3 −(3+2B)y^2 +(1−C)y]dy  B=(2−A)(1/4)−(3+2B)(1/3)+(1−C)(1/2)  ⇒3A+20B+6C=0   ...(ii)  C=∫_0 ^1 y^2 f(y)dy=∫_0 ^1 [(2−A)y^4 −(3+2B)y^3 +(1−C)y^2 ]dy  C=(2−A)(1/5)−(3+2B)(1/4)+(1−C)(1/3)  ⇒12A+30B+80C=−1   ...(iii)  from (i),(ii),(iii):  A=((71)/(453)),B=−(7/(453)),C=−((356)/(1359))  ⇒f(x)=((835x^2 )/(453))−((1345x)/(453))+((1715)/(1359))
f(x)+01(x+y)2f(y)dy=2x23x+1f(x)+01(x2+2xy+y2)f(y)dy=2x23x+1f(x)+x201f(y)dy+2x01yf(y)dy+01y2f(y)dy=2x23x+1f(x)+Ax2+2Bx+C=2x23x+1f(x)=(2A)x2(3+2B)x+(1C)A=01f(y)dy=01[(2A)y2(3+2B)y+(1C)]dyA=(2A)13(3+2B)12+(1C)8A+6B+6C=1(i)B=01yf(y)dy=01[(2A)y3(3+2B)y2+(1C)y]dyB=(2A)14(3+2B)13+(1C)123A+20B+6C=0(ii)C=01y2f(y)dy=01[(2A)y4(3+2B)y3+(1C)y2]dyC=(2A)15(3+2B)14+(1C)1312A+30B+80C=1(iii)from(i),(ii),(iii):A=71453,B=7453,C=3561359f(x)=835x24531345x453+17151359

Leave a Reply

Your email address will not be published. Required fields are marked *