anonymous
  • anonymous
Differential Equations again!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
SOLVE \[\frac{ dy }{ dx } + y = xy^{3}\]
UnkleRhaukus
  • UnkleRhaukus
you have a Bernoulli equation \[y'+p(x)y=q(x)y^n\], subsitute \(z=y^{1-n}\)
anonymous
  • anonymous
Ok..., thank u @UnkleRhaukus So.., \[\frac{ dy }{ dx } + y = xy^{3}\] \(y^{-3} \frac{ dy }{ dx } +y^{-2} = x\) ........(1) \(z = y^{1-n} \) where n = 3, then \(z = y^{-2}\) \[\frac{ dz }{dx } = \frac{ dz }{ dy }\frac{ dy }{ dx } = \frac{ d }{ dy }(y^{-2}) \frac{ dy }{ dx } = -2 y^{-3} \frac{ dy }{ dx }\] \[\frac{ dy }{ dx } = \frac{ 1 }{ -2y } y^{3} \frac{ dz }{ dx }\] substitute dy/dx into the (1) equations, becoms \[y^{-3} (\frac{ 1 }{ -2 }) y^{3} \frac{ dz }{ dx } + y^{-2} = x\] \(\frac{ dz }{ dx } -2y^{-2} = -2x\) where \(z = y^{-2}\) \(\frac{ dz }{ dx } -2z = -2x\) ....(2) i got p(x) = -2 and g(x) = -2x Then for integration factor: \[\mu(x) = \exp(\int\limits_{0}^{x} -2 dx) = \exp(-2x) = e^{-2x}\] multiply all (2) terms by \( \mu (x)\) , becomes \[e^{-2x} \frac{ dz }{ dx } -2e^{-2x}z = -2e^{-2x} x\] \[d(e^{-2x} z) = -2 e^{-2x} x\] \[e^{-2x} z = -2 \int\limits_{0}^{x} e^{-2x} x dx\] that is right??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[e^{-2x} z = \frac{ 1 }{ 2 }e^{-2x} (2x+1) + C\] where z = y^{-2} \[e^{-2x} y^{-2} = \frac{ 1 }{ 2 } e^{-2x}(2x+1) + C\]
UnkleRhaukus
  • UnkleRhaukus
thats right
UnkleRhaukus
  • UnkleRhaukus
now just rearrange and solve for y(x)
anonymous
  • anonymous
wow..., very nice.., ok., \[e^{-2x}y^{-2} = \frac{ 1 }{ 2 } e^{-2x} (2x+1) + C\] \[e^{-2x}y^{-2} - \frac{ 1 }{ 2 } e^{-2x}(2x+1) = 0\] \[y^{-2} - \frac{ 1 }{ 2 } (2x+1) = 0\] \[y^{2} = \frac{ 2 }{ (2x+1) } + C\] \[y = \sqrt{\frac{ 2 }{ (2x+1)}+ C}\] am i right????
UnkleRhaukus
  • UnkleRhaukus
almost , the term withe the constant is not quite right though , something strange happend on your second line just now ^
anonymous
  • anonymous
I'm a little bit confused about C
UnkleRhaukus
  • UnkleRhaukus
\[e^{-2x}y^{-2} - \frac{ 1 }{ 2 } e^{-2x}(2x+1) - C=0\] \[y^{-2} - \frac{ 1 }{ 2 } (2x+1) - Ce^{2x}=0\]
UnkleRhaukus
  • UnkleRhaukus
\[y^{-2}= \frac{ 1 }{ 2 }( (2x+1) + 2Ce^{2x})\] \[y^{2}= \frac{ 2 }{ (2x+1) + 2Ce^{2x}}\] \[y^{2}(x)= \frac{ 2 }{ (2x+1) + ke^{2x}}\] \[y(x)= \pm\sqrt{\frac{ 2 }{ (2x+1) + ke^{2x}}}\]
UnkleRhaukus
  • UnkleRhaukus
maybe just leave the solution as \[y^{2}(x)\]
UnkleRhaukus
  • UnkleRhaukus
\[\qquad2C=k\]
anonymous
  • anonymous
ok..., thank u so much :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.