my first Bernoulli DE ... (just looked at it on lamar tutorial and will see how will I do this one)

Question

idku · Answer

$\large \color{black}{y'=4y+e^xy^3}$

idku · Answer

(this example is completely made up, not on the tutorial)

amistre64 · Answer

how do you know it satisifies a Berny then?

amistre64 · Answer

im thinking (couild be misremembering) that  a Berny is a reduction process

idku · Answer

$\large \color{black}{y'=4y+e^xy^{3}}$
$\large \color{black}{y'-4y=e^xy^{3}}$
$\large \color{black}{y'y^{-3}-4y^{-2}=e^x}$

and then I substitute,
$\large \color{black}{v=y^{-2}}$
$\large \color{black}{v'=-2y'y^{-3}}$    --->    $\large \color{black}{(-1/2)v'=y'y^{-3}}$

my DE becomes,

$\large \color{black}{\dfrac{-1}{2}v'-4v=e^x}$

For convenience,

$\large \color{black}{v'+8v=2e^x}$

Integrating factor:   $\large \color{black}{e^{4v^2}}$

$\large \color{black}{v'e^{4v^2}+e^{4v^2}8v=e^{4v^2}2e^x}$
$\large \color{black}{dv/dx(ve^{4v^2})=e^{4v^2}2e^x}$

wait I am getting stuck here a bit

idku · Answer

how would I then integrate both sides, if I have that on the RIGHT side?

idku · Answer

oh, I missed a minus on the right side (when I multiplied by -2)

idku · Answer

$\large \color{black}{dv/dx(ve^{4v^2})=-e^{4v^2}2e^x}$

idku · Answer

the integrating facto is wrong? is it just e^x because integrating factor is with wespect to x, not v?

amistre64 · Answer

v e^(8v) = v' e^(8v) + 8v e^(8v)

amistre64 · Answer

hmm, dont think thats quite right either

idku · Answer

$\large \color{black}{v'+8v=-2e^x}$
$\large \color{black}{v'e^{8x}+e^{8x}8v=-2e^{8x}e^x}$

idku · Answer

like that?

amistre64 · Answer

that seems better yes

idku · Answer

$\large \color{black}{ve^{8x}=\int -2e^{9x}dx}$

$\large \color{black}{ve^{8x}= (-2/9)e^{9x}+c}$

$\large \color{black}{v= (-2/9)e^{x}+c/e^{8x}}$

idku · Answer

Like this?

amistre64 · Answer

so far so good

idku · Answer

$\large \color{black}{v= (-2/9)e^{x}+C/e^{8x}}$

$\large \color{black}{y^{-2}= (-2/9)e^{x}+C/e^{8x}}$

and then,

$\large \color{black}{y^2=\dfrac{1}{ (-2/9)e^{x}+C/e^{8x}}}$

$\large \color{black}{y=\pm\sqrt{\dfrac{1}{ (-2/9)e^{x}+C/e^{8x}}}}$

and the rest is a matter of algebraic manipulation if I have done it correclty.

amistre64 · Answer

the process seems fair

idku · Answer

Awesome, I prevailed it!
Thank you amistre for your confirmation!

amistre64 · Answer

youre welcome, i didnt scrutinize it, but it seemed to be in good form overall.

idku · Answer

Yes, fine for the first time. (Especially considering the fact that I am not really a "math guy")

idku · Answer

Thank you once again.