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anonymous
 5 years ago
I need help to solve this
dy/dx=xy+x2y2 // y(0)=2
thank you
anonymous
 5 years ago
I need help to solve this dy/dx=xy+x2y2 // y(0)=2 thank you

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi tonijv, This is a separable equation. Factorize the righthand side to see,\[y(x2)+(x2)=(y+1)(x2)\]You then have,\[\frac{dy}{y+1}=(x2)dx \rightarrow \int\limits_{}^{} \frac{dy}{y+1}=\int\limits_{}^{}(x2)dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ln(y+1)=\frac{x^2}{2}2x+c \rightarrow y=e^{\frac{x^2}{2}2x+c}1\]That is,\[y=Ae^{\frac{x^2}{2}2x}1\]Using the boundary condition,\[y(0)=Ae^01=2\rightarrow A=3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your equation is then,\[y=3e^{\frac{x^2}{2}2x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The previous comment was correct. This is a separable equation and wanted to also note a slightly different way of factorizing the RHS: \[dy/dx = xy + x  2y  2 = x(y + 1)  2(y+1) = (y+1) (x2)\] Now place all of the y variables to one side and all of the x variables to the other and you obtain the integral that lokisan noted above. Upon solving this integral you should get: \[\ln \left y+1 \right=x ^{2}/2  2x + C\] Now plug in the initial value conditions for x = 0 and y = 1 and solve for C to obtain: \[C=\ln(2)\] Thereby, making your final answer: \[ \ln \left y+1 \right=x ^{2}/2  2x + \ln(2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks guys is my first time here. I am lost in differentials. greetings from Barcelona

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have given you the answer "implicitly" in terms of y but you can also solve this "explicitly" for y, but some professors don't mind which form you use.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=(1/2)(x4)(x)(y+1)+2\] The derivative of the right side of the above with respect to x is: \[(1/2)(x4)(y+1)+(1/2) (x)(y+1)\] or \[x y+x2 y2\]
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