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Vectors P3

Yet another question from me! Sorry, have just finished a 80-question review exercise but there are a few I'm stuck on! Can anyone help me with the following:

Find the solution of the differential equation:
dy/dx = y/(1+x)(3+x)
given that y=2 at x=1. Express your answer in the form y^2=f(x)

In the earlier part of the question I worked out 1/(1+x)(3+x) in partial fractions to be 1/2(1+x) - 1/2(3+x). So I assumed that to solve that differential eqtn i used:

Int 1/y dy= Int x. dx

And continued to solve it but I don't get the right answer of y^2 = 8(1+x)/(3+x)

Thanks :smile:

Reply 1

mikesgt2

The solution is given by:

INT 1/y dy = INT 1/(1+x)(3+x) dx

You are right about the partial fractions, so the solution is then

INT 1/y dy = (1/2) INT 1/(1+x) - 1/(3+x) dx
ln|y| = (1/2)[ ln|1+x| - ln|3+x| + lnA ] (write the constant as lnA rather than C)

So, mutiplying by 2 and combing the logarithms on the right gives

2ln|y| = ln| A(1+x)/(3+x) |

Note that 2ln|y| = ln|y^2|, therefore

ln|y^2| = ln| A(1+x)/(3+x) |
y^2 = A(1+x)/(3+x)

Sub y=2,x=1:

4 = A(2/4) => A=8

Which gives the result

y^2 = 8(1+x)/(3+x)

I hope this helps.