Find the integral of ((2(7x^(2)-xe^(-2x))-5)/x) . Given that y=27 at x=1, solve the differential equation dy/dx=((2(7x^(2)-xe^(-2x))-5)/-3x).y^(2/3) in terms of y.

Part A)Expand numerator in the integral to get 14x^(2) - 2xe^(-2x)-5Now divide by denominator to get 14x - 2e^(-2x) - 5/xNow integrate to get 7x^(2) + e^(-2x) - 5ln(x)Part B)Get all components of y on the left, and all components of x on the right to get -3⌠ y^(-2/3) dy = ⌠ ((2(7x^(2)-xe^(-2x))-5)/x) dx ⌡ ⌡ 3. solve to get -9y^(1/3) = 7x^(2) + e^(-2x) - 5ln(x) + C 4. Now substitute given values of x and y (1 and 27 respectively), in order to calculate C -27 = 7 + e^(-2) - 0 + c C = -34-e^(-2) 5. Substituting in C we get -9y^(1/3) = 7x^(2) + e^(-2x) - 5ln(x) + (-34-e^(-2)) 6. Rearranging in terms of y we get y = (-1/729).(7x^(2) + e^(-2x) - 5ln(x) - 34 - e^(-2))^(3)

GR
Answered by George R. Maths tutor

4025 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Why is the definite integral between negative limits of a function with positive values negative even though the area bound by the x-axis is positive? for example the integral of y=x^2 between x=-2 and x=-1


Find the exact value of the integral of (2+7/x), between x=1 and x=e. Give your answer in terms of e.


Sketch 20x--x^2-2x^3


Prove that, if 1 + 3x^2 + x^3 < (1+x)^3, then x>0


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences