Answered: Solve the initial value problem. dy 6…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please help solve the image below. I separated each variable and integrated both sides beginning with x^6 dx = 1/(y-5) dy. After integration I got (x^7)/7 - ln|y-5| = C, and solved for C using the initial values provided. I then tried to get the implicit solution, but got tripped up. I think I messed up when I took 'e' of both sides to get rid of the 'ln'. Thank you for your help and please hand write clearly.

Solve the initial value problem.
dy
|$
dx
6
= x(y-5), y(0) = 8
The solution is
(Type an implicit solution. Type an equation using x and y as the variables.)

Expert Solution

Step 1: ''Introduction to the solution''

The given differential equation is $fraction numerator d y over denominator d x end fraction equals x to the power of 6 open parentheses y minus 5 close parentheses....... left parenthesis 1 right parenthesis comma y left parenthesis 0 right parenthesis equals 8$

We have to solve the given differential equation(1).

Now,from(1), we have, $fraction numerator d y over denominator left parenthesis y minus 5 right parenthesis end fraction equals x to the power of 6 space d x$

$rightwards double arrow fraction numerator d left parenthesis y minus 5 right parenthesis over denominator open parentheses y minus 5 close parentheses end fraction equals x to the power of 6 d x$

Integrating with respect to'x', we get,

$ln left parenthesis y minus 5 right parenthesis$ $equals x to the power of 7 over 7 plus log left parenthesis C right parenthesis$ ,where $C$ being constant.

$rightwards double arrow log open parentheses fraction numerator y minus 5 over denominator C end fraction close parentheses equals x to the power of 7 over 7......... left parenthesis 2 right parenthesis$