Problem in defferential equation

• Jan 25th 2008, 07:16 PM
geton
Problem in defferential equation
1. Find the general solution of linear differential equation

dy/dx – y/(x+1) = x

2. Given that (x) dy/dx – 2y = x^3 ln x, find y in terms of x such that y=2 at x=1.

• Jan 25th 2008, 07:50 PM
mr fantastic
Quote:

Originally Posted by geton
1. Find the general solution of linear differential equation

dy/dx – y/(x+1) = x

2. Given that (x) dy/dx – 2y = x^3 ln x, find y in terms of x such that y=2 at x=1.

1. Use the integrating factor method. I.F. = $\displaystyle e^{-\int \frac{1}{x+1} \, dx} = e^{-\ln|x + 1|} = e^{\ln \frac{1}{|x + 1|}} = \frac{1}{x + 1}$.

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2. Divide through by x. Then use the integrating factor method. I.F. = $\displaystyle e^{-2 \int \frac{1}{x} \, dx} = e^{-2 \ln|x|} = e^{\ln \frac{1}{x^2}} = \frac{1}{x^2}$.
• Jan 25th 2008, 08:12 PM
geton
Quote:

Originally Posted by mr fantastic
1. Use the integrating factor method. I.F. = $\displaystyle e^{-\int \frac{1}{x+1} \, dx} = e^{-\ln|x + 1|} = e^{\ln \frac{1}{|x + 1|}} = \frac{1}{x + 1}$.

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2. Divide through by x. Then use the integrating factor method. I.F. = $\displaystyle e^{-2 \int \frac{1}{x} \, dx} = e^{-2 \ln|x|} = e^{\ln \frac{1}{x^2}} = \frac{1}{x^2}$.

1. Multiplying by (x+1)^-1 gives:

(x+1)^-1 dy/dx - y (x+1)^-2 = x(x+1)^-1

y(x+1)^1 d/dx = ∫ x(x+1)^-1

How can I integrated this?

2. After Integrating:

yx^-2 = x(ln x - 1) + C

Since y=2 at x=1 this gives:

C=3

That is: yx^-2 = x(ln x - 1) + 3

So y =x^3 (ln x - 1) + 3 x^2

Is it right?
• Jan 25th 2008, 09:14 PM
mr fantastic
Quote:

Originally Posted by geton
[snip]
∫ x(x+1)^-1

How can I integrated this?
[snip]

$\frac{x}{x + 1} = \frac{(x + 1) - 1}{x + 1} = 1 - \frac{1}{x + 1}$
• Jan 25th 2008, 09:37 PM
geton
Thank you mr fantastic. But is my 2nd problem's solution right?
• Jan 25th 2008, 09:57 PM
geton
Yes I found my mistake on 2nd problem :)

Problem resolved.
• Jan 26th 2008, 12:13 AM
mr fantastic
Quote:

Originally Posted by geton
Yes I found my mistake on 2nd problem :)

Problem resolved.

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