Question

Solve this DE by seperation of variables: yln(x)x'=(y+1)^2/x^2

Answer 1

alright i split it up to get

\[yln(x)\frac{dx}{dy}=\frac{(y+1)^2}{x^2}\]

Answer 2

then made a common factor to fact out

CF=\[\frac{x^2}{y}dy\]

which got me to

\[x^2ln(x)dx=\frac{(y+1)^2}{y}dy\]

Answer 3

squaring the top and seperating the numerator while also taking anti derivative

\[\int\limits_{}^{}x^2\ln(x)dx=\int\limits_{}^{}\frac{y^2}{y}dy+\int\limits_{}^{}\frac{2y}{y}dy+\int\limits_{}^{}\frac{dy}{y}\]

Answer 4

simplifying and doing the right side you get

\[\int\limits_{}^{}x^2\ln(x)dx=\frac{y^2}{2}+2y+\ln(y)+c\]

Answer 5

now doing the right side by parts is where i got something different from the book

Answer 6

using by parts i get

\[u=ln(x)\]
\[du=\frac{1}{x}\]

Answer 7

\[dv=x^2\]
\[v=\frac{x^3}{3}\]

Answer 8

\[uv-\int\limits_{}^{}vdu=\frac{x^3}{3}\ln(x)-\int\limits_{}{}\frac{x^2}{x}dx\]

Answer 9

]

Answer 10

not sure what happened there sam as i can't see anything

Answer 11

simplifying the integral and doing it yields

\[\frac{x^3}{3}\ln(x)-\frac{x^2}{2}+c_2\]

Answer 12

but that is not what is within the book?

Answer 13

trying to notice my mistake

Answer 14

or if it's a mistake in the book

Answer 15

the y side is correct however this is what they have for the answer

\[\frac{x^3}{3}\ln(x)-\frac{x^3}{9}=\frac{y^2}{2}+2y+\ln(y)+c\]

Answer 16

Sorry I thought you are doing it by using integrating factor:(

Answer 17

i found the error instead of x^3 i had x^2 only idk why but

thanks anyway

Answer 18

somewhere inbetween i must have mixed it up

Answer 19

\[\int\limits_{}{}\frac{x^3}{3x}dx=\frac{1}{3}\int\limits_{}{}x^2dx=\frac{1}{9}x^3\]

Answer 20

Seems right:)

Answer 21

\[\frac{1}{3}x^3(\ln(x)-\frac{1}{3})\]