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Given that $y = 35$ at $x = 4$, find $y$ in terms of $x$, giving each term in its simplest form. - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 2

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Given-that-$y-=-35$-at-$x-=-4$,-find-$y$-in-terms-of-$x$,-giving-each-term-in-its-simplest-form.-Edexcel-A-Level Maths: Pure-Question 6-2010-Paper 2.png

Given that $y = 35$ at $x = 4$, find $y$ in terms of $x$, giving each term in its simplest form.

Worked Solution & Example Answer:Given that $y = 35$ at $x = 4$, find $y$ in terms of $x$, giving each term in its simplest form. - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 2

Step 1

Step 1: Solve the Differential Equation

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Answer

We start with the differential equation:

dydx=5x12+xx,x>0\frac{dy}{dx} = 5x^{\frac{1}{2}} + x\sqrt{x}, \quad x > 0

To solve this, we will integrate both sides with respect to xx:

y=∫(5x12+xx)dxy = \int \left( 5x^{\frac{1}{2}} + x\sqrt{x} \right) dx

The term xxx\sqrt{x} can be rewritten as x32x^{\frac{3}{2}}. Therefore,

y=∫(5x12+x32)dxy = \int \left( 5x^{\frac{1}{2}} + x^{\frac{3}{2}} \right) dx

This results in:

y=(5â‹…23x32+25x52)+C=103x32+25x52+Cy = \left( 5 \cdot \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} \right) + C = \frac{10}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

Step 2

Step 2: Use the Initial Condition

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Answer

We know that y=35y = 35 when x=4x = 4:

35=103(4)32+25(4)52+C35 = \frac{10}{3} (4)^{\frac{3}{2}} + \frac{2}{5} (4)^{\frac{5}{2}} + C

Calculating the terms:

  1. 432=84^{\frac{3}{2}} = 8 thus:

35=103â‹…8+25â‹…32+C35 = \frac{10}{3} \cdot 8 + \frac{2}{5} \cdot 32 + C

  1. 103â‹…8=803\frac{10}{3} \cdot 8 = \frac{80}{3} and 25â‹…32=645\frac{2}{5} \cdot 32 = \frac{64}{5}

  2. Converting to a common denominator of 15 gives:

803=40015\frac{80}{3} = \frac{400}{15} 645=19215\frac{64}{5} = \frac{192}{15}

So we have:

35=40015+19215+C35 = \frac{400}{15} + \frac{192}{15} + C

Thus:

C=35−59215=525−59215=−6715C = 35 - \frac{592}{15} = \frac{525 - 592}{15} = \frac{-67}{15}

Step 3

Step 3: Write the Final Result for y

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Answer

Substituting CC back into the equation gives:

y=103x32+25x52−6715y = \frac{10}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} - \frac{67}{15}

This is the required expression for yy in terms of xx, with each term in its simplest form.

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