A question definesf(x)=(ax+b)exand states that it satisfies the equationf(x)=∫0xex−yf′(y)dy−(x2−x+1)exYou're required to find a and b.

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Krystian Quintero

2022-03-01

A question defines

f (x) = (a x + b) e^{x}

and states that it satisfies the equation

f (x) = \int_{0}^{x} e^{x - y} f^{'} (y) d y - (x^{2} - x + 1) e^{x}

You're required to find a and b.

Capodarcod0f

Beginner2022-03-02Added 6 answers

Step 1
Going by your method, we get,

f^{'} (x) = e^{x} (a x + b + a)

\Rightarrow f (x) = \int_{0}^{x} e^{x - y} e^{y} (a y + b + a) d y - (x^{2} - x + 1) e^{x}

\Rightarrow f (x) = e^{x} \int_{0}^{x} (a y + b + a) d y - (x^{2} - x + 1) e^{x}

\Rightarrow f (x) = e^{x} (d \frac{a x^{2}}{2} + (b + a) x) - (x^{2} - x + 1) e^{x}

\Rightarrow (a x + b) e^{x} = e^{x} ((d \frac{a}{2} - 1) x^{2} + (b + a + 1) x - 1)

Cancelling

e^{x}

and equating coefficient both sides, you get

a = 2

and

b = - 1

.
Going by their method, we get,

e^{- x} f (x) = \int_{0}^{x} e^{- y} f^{'} (y) d y - (x^{2} - x + 1)

\Rightarrow d \frac{d}{d x} (a x + b) = e^{- x} f^{'} (x) - (2 x - 1)

\Rightarrow a = (a x + b + a) - (2 x - 1)

\Rightarrow 0 = (a - 2) x + (b + 1) = 0

which leads to the same solution. Note that the given answer that

b = 1

is wrong.

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