Question

Find the particular solution for 2x (dy/dx)-ln x^2 that satisfies the initial condition y(1) = 2. Explain your steps.

Answer 1

\[2x \frac{ dy }{ dx }-\ln x ^{2}\]

equation looks like that

Answer 2

separation of variables

Answer 3

ok i need help with the whole problem

Answer 4

sorry was afk...

Answer 5

\(\large 2x \frac{ dy }{ dx }-\ln x^{2}\)

\(\large 2x \frac{ dy }{ dx } = \ln x^{2}\)

\(\large 2 ~dy = \frac{\ln x^{2}}{x}dx \)

Answer 6

Integrate both sides

Answer 7

\(\large \int 2 ~dy = \int \frac{\ln x^{2}}{x}dx \)

Answer 8

evaluate

Answer 9

ok thank you!

Answer 10

u sure.. u can manage the rest ha ? :)

Answer 11

integrating right side can be tricky...
u need to use "u sub"

Answer 12

no i cant hahah

Answer 13

\(\large \int 2 ~dy = \int \frac{\ln x^{2}}{x}dx \)

\(\large 2\int ~dy = \int \frac{2\ln x}{x}dx \)

\(\large 2y = 2\int \frac{\ln x}{x}dx \)

Answer 14

is that it?

Answer 15

substitute \(\ln x = t\)
\(\frac{1}{x} dx = dt\)

right side beocmes :

\(\large 2\int t~ dt \)

\(\large 2 \frac{t^2}{2} + C \)

\(\large t^2+ C \)

\(\large (\ln x)^2+ C \)

Answer 16

plug that in right hand side

Answer 17

\(\large 2y = (\ln x)^2 + C\)

Answer 18

you need to plugin the initial condition (1, 2) and solve \(C\) :

\(\large 2*2 = (\ln 1)^2 + C \implies C = 4\)

Answer 19

So, the solution is :

\(\large 2y = (\ln x)^2 + 4\)

Answer 20

thank you so much for your help! This'll definitely help me with similar problems

Answer 21

np :) good luck !

Answer 22

i made a mistake earlier

Answer 23

nvm, it is correct :)