Solve the initial-value problem using the method of undetermined coefficients. y'' − y' = xe^x, y(0) = 6, y'(0) = 5

First check the characteristic solution. The characteristic equation to this DE is

r ² - r = r (r - 1) = 0

with roots r = 0 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(0x) + C₂ exp(1x)

y (char.) = C₁ + C₂ exp(x)

For the particular solution, we try the ansatz

y (part.) = (ax + b) exp(x)

but exp(x) is already accounted for in the second term of y (char.), so we multiply each term here by x :

y (part.) = (ax ² + bx) exp(x)

Differentiate this twice and substitute the derivatives into the DE.

y' (part.) = (2ax + b) exp(x) + (ax ² + bx) exp(x)

… = (ax ² + (2a + b)x + b) exp(x)

y'' (part.) = (2ax + 2a + b) exp(x) + (ax ² + (2a + b)x + b) exp(x)

… = (ax ² + (4a + b)x + 2a + 2b) exp(x)

(ax ² + (4a + b)x + 2a + 2b) exp(x) - (ax ² + (2a + b)x + b) exp(x)

= x exp(x)

The factor of exp(x) on both sides is never zero, so we can cancel them:

(ax ² + (4a + b)x + 2a + 2b) - (ax ² + (2a + b)x + b) = x

Collect all the terms on the left side to reduce it to

2ax + 2a + b = x

Matching coefficients gives the system

2a = 1

2a + b = 0

and solving this yields

a = 1/2, b = -1

Then the general solution to this DE is

y(x) = C₁ + C₂ exp(x) + (1/2 x ² - x) exp(x)

For the given initial conditions, we have

y (0) = C₁ + C₂ = 6

y' (0) = C₂ - 1 = 5

and solving for the constants here gives

C₁ = 0, C₂ = 6

so that the particular solution to the IVP is

y(x) = 6 exp(x) + (1/2 x ² - x) exp(x)

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step-by-step explanation:

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.