Question

help with diff eqns

Answer 1

\[y ^{2}y'+2xy ^{3}=6x\]

Answer 2

put
\[y^3=t,3 y^2\frac{ dy }{ dx }=\frac{ dt }{ dx },y^2\frac{ dy }{ dx }=\frac{ 1 }{ 3 }\frac{ dt }{ dx }\]
\[\frac{ 1 }{ 3 }\frac{ dt }{ dx }+3xt=6x\]

Answer 3

correction write 2xt in place of 3xt
\[\frac{ dt }{ dx }+6xt=18x\]

Answer 4

\[I.F=e ^{\int\limits6x dx}=e ^{\frac{ 6x^2 }{ 2 }}=e ^{3x^2}\]

Answer 5

c.s is
\[t.e ^{3 x^2}=\int\limits (18x)e ^{3x^2}dx+c\]
put \[3 x^2=z,6 x ~dx=dz\]
complete it.

Answer 6

gimme a few min please

Answer 7

ok, im not following

Answer 8

=\[3e ^{3x ^{2}}\]
?

Answer 9

ok I'm not the best with DE, but I'll give it a shot

Answer 10

what surjithayer did was say

let y^3 = t

to replace something complicated with something simpler

Answer 11

y is a function of x

so if y^3 = t, then t must also be a function of x (since t is taking over 'y's job so to speak)

Answer 12

thanks

Answer 13

surjithayer then derived both sides of y^3 = t to get

3y^2*(dy/dx) = dt/dx

surjithayer derived with respect to x (notice the chain rule on the left side)

Answer 14

doing this isn't clear until you see the y^2 and the dy/dx

that's in the original equation as well, so let's isolate that

3y^2*(dy/dx) = dt/dx

y^2*(dy/dx) = (1/3)*(dt/dx)