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f-x-f-x-0-1-f-x-dx-f-0-1-f-x-

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Question Number 163119 by tounghoungko last updated on 04/Jan/22
f ′(x)= f(x)+∫_0 ^1 f(x)dx f(0)=1 ⇒f(x)=?
f′(x)=f(x)+∫01f(x)dxf(0)=1⇒f(x)=?
Answered by mr W last updated on 04/Jan/22
y′=y+k with k=∫_0 ^1 f(x)dx (dy/(y+k))=dx ∫(dy/(y+k))=∫dx ln (y+k)=x+C_1 y+k=Ce^x y=f(x)=Ce^x −k ∫_0 ^1 f(x)dx=[Ce^x −kx]_0 ^1 =C(e−1)−k=^! k ⇒C=((2k)/(e−1)) f(x)=(((2e^x )/(e−1))−1)k f(0)=((2/(e−1))−1)k=^! 1 ⇒k=((e−1)/(3−e)) ⇒f(x)=(((e−1)/(3−e)))(((2e^x )/(e−1))−1)
y′=y+kwithk=∫01f(x)dxdyy+k=dx∫dyy+k=∫dxln(y+k)=x+C1y+k=Cexy=f(x)=Cex−k∫01f(x)dx=[Cex−kx]01=C(e−1)−k=!k⇒C=2ke−1f(x)=(2exe−1−1)kf(0)=(2e−1−1)k=!1⇒k=e−13−e⇒f(x)=(e−13−e)(2exe−1−1)
Commented by Tawa11 last updated on 04/Jan/22
Great sir
Greatsir
Answered by tounghoungko last updated on 04/Jan/22

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