Given that \( \frac{dy}{dx} = -x^3 + \frac{4x - 5}{2x^2} \), \ x \neq 0 \n
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 10 - 2013 - Paper 3
Question 10
Given that \( \frac{dy}{dx} = -x^3 + \frac{4x - 5}{2x^2} \), \ x \neq 0 \n
Given that \( y = 7 \) at \( x = 1 \), find \( y \) in terms of \( x \), giving each term... show full transcript
Worked Solution & Example Answer:Given that \( \frac{dy}{dx} = -x^3 + \frac{4x - 5}{2x^2} \), \ x \neq 0 \n
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 10 - 2013 - Paper 3
Step 1
1. Integrate the Differential Equation
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find ( y ) in terms of ( x ), we start by integrating ( \frac{dy}{dx} ):
∫dy=∫(−x3+2x24x−5)dx
Now, we can simplify the right-hand side:
=∫−x3dx+∫(2x4−2x25)dx
Calculating each integral gives:
=−4x4+2ln∣x∣+2x5+C
Step 2
2. Apply Initial Condition
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Next, we use the initial condition ( y = 7 ) when ( x = 1 ) to find the constant ( C ).
Substituting these values into the integrated equation:
7=−414+2ln∣1∣+2⋅15+C
Simplifying this gives:
7=−41+0+25+C
Solving for ( C ):
7=−41+410+C⇒C=7−49=428−49=419
Step 3
3. Final Expression for \( y \)
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Substituting the value of ( C ) back into the equation gives us:
y=−4x4+2ln∣x∣+2x5+419
This is the expression for ( y ) in terms of ( x ), with each term in its simplest form.
