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.
Please help me to solve these.
1. Use the integrating factor method. I.F. = $\displaystyle \displaystyle e^{-\int \frac{1}{x+1} \, dx} = e^{-\ln|x + 1|} = e^{\ln \frac{1}{|x + 1|}} = \frac{1}{x + 1}$.
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2. Divide through by x. Then use the integrating factor method. I.F. = $\displaystyle \displaystyle e^{-2 \int \frac{1}{x} \, dx} = e^{-2 \ln|x|} = e^{\ln \frac{1}{x^2}} = \frac{1}{x^2}$.
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?