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natka813 [3]
3 years ago
10

Max's Moving Company charges a $75 loading fee to move your boxes. They also charge $2 per mile that they have to drive the van

to move you boxes from one place to another.
Mathematics
1 answer:
maks197457 [2]3 years ago
3 0

Answer:

y=75+2x

Step-by-step explanation:

Given data

Loading fee=$75

Charge per mile= $2

Let the distance be x

and the total charge be y

Hence the expression for the total charge is

y=75+2x

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Solve the initial value problems.
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Both equations are linear, so I'll use the integrating factor method.

The first ODE

xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0

has integrating factor

\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x

In the original equation, multiply both sides by eˣ :

xe^x y' + (x+1) e^x y = 0

Observe that

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so that the left side is the derivative of a product, namely

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Integrate both sides with respect to x :

\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx

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Solve for y :

y = \dfrac{C}{xe^x}

Use the given initial condition to solve for C. When x = 1, y = 2, so

2 = \dfrac{C}{1\cdot e^1} \implies C = 2e

Then the particular solution is

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The second ODE

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has integrating factor

\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}

Multiply both sides of the equation by 1/(1 + x²) :

\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

\left(\dfrac1{1+x^2}y\right)' = 0

\dfrac1{1+x^2}y = C

y = C(1 + x^2)

When x = 0, y = 3, so

3 = C(1+0^2) \implies C=3

\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}

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