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gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2SOLVE \[\frac{ dy }{ dx } + y = xy^{3}\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1you have a Bernoulli equation \[y'+p(x)y=q(x)y^n\], subsitute \(z=y^{1n}\)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2Ok..., thank u @UnkleRhaukus So.., \[\frac{ dy }{ dx } + y = xy^{3}\] \(y^{3} \frac{ dy }{ dx } +y^{2} = x\) ........(1) \(z = y^{1n} \) 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??

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2\[e^{2x} z = \frac{ 1 }{ 2 }e^{2x} (2x+1) + C\] where z = y^{2} \[e^{2x} y^{2} = \frac{ 1 }{ 2 } e^{2x}(2x+1) + C\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1now just rearrange and solve for y(x)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2wow..., 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
 2 years ago
Best ResponseYou've already chosen the best response.1almost , the term withe the constant is not quite right though , something strange happend on your second line just now ^

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2I'm a little bit confused about C

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[e^{2x}y^{2}  \frac{ 1 }{ 2 } e^{2x}(2x+1)  C=0\] \[y^{2}  \frac{ 1 }{ 2 } (2x+1)  Ce^{2x}=0\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[y^{2}= \frac{ 1 }{ 2 }( (2x+1) + 2Ce^{2x})\] \[y^{2}= \frac{ 2 }{ (2x+1) + 2Ce^{2x}}\] \[y^{2}(x)= \frac{ 2 }{ (2x+1) + ke^{2x}}\] \[y(x)= \pm\sqrt{\frac{ 2 }{ (2x+1) + ke^{2x}}}\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1maybe just leave the solution as \[y^{2}(x)\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.2ok..., thank u so much :)
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