Explanation Given: An integral equation, x ( t ) = 3 + 4 ∫ 0 t x ( t − τ ) cos τ d τ Approach: Consider Volterra integral equation, x ( t ) = f ( t ) + ∫ 0 t k ( t − τ ) x ( τ ) d τ ( 1 ) Here, x ( t ) is an unknown function, k is the kernel, ∫ 0 t k ( t − τ ) x ( τ ) d τ is a convolution integral. Laplace transform of equation ( 1 ) yields, X ( s ) = F ( s ) + L [ k ( t ) ∗ x ( t ) ] Use the convolution theorem for Laplace transforms. For continuous functions f and g in E ( 0 , ∞ ) , L [ f ∗ g ] = L [ f ] L [ g ] is the convolution theorem. X ( s ) = F ( s ) + K ( s ) X ( s ) Solve for X ( s ) , X ( s ) = F ( s ) 1 − K ( s ) Take inverse Laplace on both the sides to solve for x ( t ) , x ( t ) = L − 1 [ F ( s ) 1 − K ( s ) ] Use the properties and table ( 1 ) of Laplace transforms to determine x ( t ) . The Laplace transform satisfies the linearity properties L [ f + g ] = L [ f ] + L [ g ] L [ c f ] = c L [ f ] For all transformable functions f and g and constants c . Summary of Laplace Transforms Function f ( t ) Laplace Transform F ( s ) f ( t ) = t n , n is a nonnegative integer F ( s ) = n ! s n + 1 , s > 0 . f ( t ) = e a t , a constant F ( s ) = 1 s − a , s > a . f ( t ) = sin b t , b constant F ( s ) = b s 2 + b 2 , s > 0 . f ( t ) = cos b t , b constant F ( s ) = s s 2 + b 2 , s > 0 . f ( t ) = t − 1 / 2 F ( s ) = ( π / s ) 1 / 2 , s > 0 . f ( t ) = u a ( t ) (see Section 10.7) F ( s ) = 1 s e − a s . f ( t ) = δ ( t − a ) (see Section 10.8) F ( s ) = e − a s . Transform of Derivatives (see Section 10.4) f ′ L [ f ′ ] = s L [ f ] − f ( 0 ) . f ′ ′ L [ f ′ ′ ] = s 2 L [ f ] − s f ( 0 ) − f ′ ( 0 ) . Shifting theorems (see Section 10.5 and 10.7) e a t f ( t ) F ( s − a ) . u a ( t ) f ( t − a ) . e − a s F ( s ) . Table ( 1 ) Calculation: Given Volterra integral equation, x ( t ) = 3 + 4 ∫ 0 t x ( t − τ ) cos τ d τ Substitute w = t − τ in the given integral equation,

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Solution Summary: The author explains that the solution to the given integral equation is x(t)=3+2sqrt

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