Find the curve whose gradient is given by dy/dx=xy and which passes through the point (0,3)

First "Separate the Variables" by rearranging the equation to get the ys on the LHS and the xs on the RHS:

(1/y) dy=x dx

Now Integrate:

Integral(1/y) dy = Integral(x) dx

ln(y)=x2/2 + constant of integration (c)

Rearrange to get y=:

e(lny)=e(x2/2)+c

y=e(x^2/2)+c = e* ex^2/2 = Ae0.5x^2

This is your GENERAL SOLUTION (GS)

Now plug in the coordinates:

3=Ae0.50=A1=A

A=3

So:

y=3e0.5x^2

This is the PARTICUAR SOLUTION (PS) and also the answer to original question

CC
Answered by Christian C. Maths tutor

3481 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

An ellipse has the equation (x^2)/4 + (y^2)/9 = 1. Find the equation of the tangent at (-6/5 , 12/5)


Differentiate the equation y = x^2 + 3x + 1 with respect to x.


June 2008 C1 Paper Differentiation Question


How do i use chain rule to calculate the derivative dy/dx of a curve given by 2 "parametric equations": x=(t-1)^3, y=3t-8/t^2


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