Let phi (x) = int_{0}^{1} e^{x} e^{t} phi (t) | vedclass

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(A) Given $\phi(x) = e^{x} \int_{0}^{1} e^{t} \phi(t) dt + x$. Let $C = \int_{0}^{1} e^{t} \phi(t) dt$. Then $\phi(x) = C e^{x} + x$.Substituting $\ph

Vedclass · Accepted Answer

$A = \ln(e^{2} - 3) - 2$

Vedclass · Answer

(A) Given $\phi(x) = e^{x} \int_{0}^{1} e^{t} \phi(t) dt + x$. Let $C = \int_{0}^{1} e^{t} \phi(t) dt$. Then $\phi(x) = C e^{x} + x$.Substituting $\phi(t) = C e^{t} + t$ into the integral:$C = \int_{0}^{1} e^{t} (C e^{t} + t) dt = \int_{0}^{1} (C e^{2t} + t e^{t}) dt$.Evaluating the integral:$C = C \left[ \frac{e^{2t}}{2} \right]_{0}^{1} + \left[ t e^{t} - e^{t} \right]_{0}^{1} = C \left( \frac{e^{2} - 1}{2} \right) + (e - e) - (0 - 1) = C \left( \frac{e^{2} - 1}{2} \right) + 1$.$C \left( 1 - \frac{e^{2} - 1}{2} \right) = 1 \Rightarrow C \left( \frac{2 - e^{2} + 1}{2} \right) = 1 \Rightarrow C \left( \frac{3 - e^{2}}{2} \right) = 1 \Rightarrow C = \frac{2}{3 - e^{2}}$.Thus,$\phi(x) = \frac{2}{3 - e^{2}} e^{x} + x$.Now,$\phi(\ln(e^{2} - 3)) = \frac{2}{3 - e^{2}} e^{\ln(e^{2} - 3)} + \ln(e^{2} - 3) = \frac{2}{3 - e^{2}} (e^{2} - 3) + \ln(e^{2} - 3) = -2 + \ln(e^{2} - 3)$.

Let $\phi (x) = \int_{0}^{1} e^{x} e^{t} \phi (t) dt + x$. If $\phi (\ln (e^{2} - 3))$ is equal to $A$,then find the value of $A$.

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