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Given that y = 7 at x = 1, find y in terms of x, giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 3

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Given that y = 7 at x = 1, find y in terms of x, giving each term in its simplest form. $$\frac{dy}{dx} = -x^3 + \frac{4x - 5}{2x^2}, \quad x \neq 0$$

Worked Solution & Example Answer:Given that y = 7 at x = 1, find y in terms of x, giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 3

Step 1

Find the Indefinite Integral

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Answer

We have the equation:

dydx=x3+4x52x2\frac{dy}{dx} = -x^3 + \frac{4x - 5}{2x^2}

To proceed, we first simplify the right-hand side:

dydx=x3+42x152x2\frac{dy}{dx} = -x^3 + \frac{4}{2}x^{-1} - \frac{5}{2}x^{-2}

which simplifies to:

dydx=x3+2x152x2\frac{dy}{dx} = -x^3 + 2x^{-1} - \frac{5}{2}x^{-2}

Next, we integrate each term with respect to x:

y=(x3+2x152x2)dxy = \int (-x^3 + 2x^{-1} - \frac{5}{2}x^{-2}) \, dx

Step 2

Perform the Integration

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Answer

The integral can be computed as follows:

y=14x4+2lnx+521x+cy = -\frac{1}{4}x^4 + 2\ln|x| + \frac{5}{2} \cdot \frac{1}{x} + c

Thus,

y=14x4+2lnx+52x1+cy = -\frac{1}{4}x^4 + 2\ln|x| + \frac{5}{2}x^{-1} + c

Step 3

Apply the Initial Condition

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Answer

Now we apply the initial condition:

y = 7 at x = 1.
Substituting these values into the equation gives:

7=14(1)4+2ln(1)+52(1)+c7 = -\frac{1}{4}(1)^4 + 2\ln(1) + \frac{5}{2}(1) + c

Since ln(1) = 0,
we simplify to:

7=14+52+c7 = -\frac{1}{4} + \frac{5}{2} + c

Step 4

Solve for c

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Answer

This simplifies to
7=14+104+c7 = -\frac{1}{4} + \frac{10}{4} + c

So,

7=94+c7 = \frac{9}{4} + c

Solving for c yields:

c=794=28494=194c = 7 - \frac{9}{4} = \frac{28}{4} - \frac{9}{4} = \frac{19}{4}

Step 5

Final Expression for y

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Answer

Substituting c back into the equation, we have:

y=14x4+2lnx+52x1+194y = -\frac{1}{4}x^4 + 2\ln|x| + \frac{5}{2}x^{-1} + \frac{19}{4}

This is the required expression for y in terms of x.

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