Question

pls help me solve xdy/dx-y(+y^2)e^2x=0 using y=v^-1

Answer 1

\[x \frac{ dy }{ dx}-y+y ^{2}e ^{2x}=0 using y=v ^{-1}\]

Answer 2

hmm, a bernoulli equation with n = 2.

Answer 3

the right side is \[-\frac{ e^{2x} }{ x }\]

Answer 4

oh well, i'll take it from here.
divide by \(xy^2\)\[\huge \frac{ y^{-2}dy }{ dx }-\frac{ 1 }{ x } \underbrace{y^{-1}}_u = -\frac{e^{2x}}{x}\]

Answer 5

\[u=y^{-1}\]\[du=-y^{-2}dy\]

\[\large -\frac{ du }{ dx }-\frac{ 1 }{ x }u=-\frac{ e^{2x} }{ x }\]

Answer 6

\[\frac{ du }{ dx }+\frac{ 1 }{ x }u=\frac{ e^{2x} }{ x }\]\[u(x)=\int\limits x \left( \frac{ e^{2x} }{ x } \right)dx \]

Answer 7

That'll give \[\frac{ 1 }{ y}=\frac{ 1 }{ 2}e ^{2x}\]

Answer 8

forgot +c

Answer 9

then \[y= \frac{ 2 }{ e ^{2x}}+c _{2}\]

Answer 10

\[u(x)=\frac{ 1 }{ y }x\], its multiplication, not function notation. apologies.

Answer 11

\[\frac{ x }{ y }=\frac{ e^{2x} }{ 2 }+C\]

Answer 12

yep, thx, was juss rechecking noticed ma mistake

Answer 13

Thanks a million

Answer 14

forgot to ask if they solve do i have to give in terms of y or i can leave it in implicit form

Answer 15

i would write the final answer free of denominators.
\[2x=y e^{2x} + 2Cy\]

Answer 16

thx

Answer 17

yw