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

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
Ok..., 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??
 one year ago

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
thats right
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
now just rearrange and solve for y(x)
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
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????
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
almost , the term withe the constant is not quite right though , something strange happend on your second line just now ^
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
I'm a little bit confused about C
 one year ago

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

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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\qquad2C=k\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
ok..., thank u so much :)
 one year ago
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