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anonymous

  • 5 years ago

I need help to solve this dy/dx=xy+x-2y-2 // y(0)=2 thank you

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  1. anonymous
    • 5 years ago
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    Hi tonijv, This is a separable equation. Factorize the right-hand side to see,\[y(x-2)+(x-2)=(y+1)(x-2)\]You then have,\[\frac{dy}{y+1}=(x-2)dx \rightarrow \int\limits_{}^{} \frac{dy}{y+1}=\int\limits_{}^{}(x-2)dx\]

  2. anonymous
    • 5 years ago
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    \[\ln(y+1)=\frac{x^2}{2}-2x+c \rightarrow y=e^{\frac{x^2}{2}-2x+c}-1\]That is,\[y=Ae^{\frac{x^2}{2}-2x}-1\]Using the boundary condition,\[y(0)=Ae^0-1=2\rightarrow A=3\]

  3. anonymous
    • 5 years ago
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    Your equation is then,\[y=3e^{\frac{x^2}{2}-2x}\]

  4. anonymous
    • 5 years ago
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    The previous comment was correct. This is a separable equation and wanted to also note a slightly different way of factorizing the RHS: \[dy/dx = xy + x - 2y - 2 = x(y + 1) - 2(y+1) = (y+1) (x-2)\] 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: \[ \ln \left| y+1 \right|=x ^{2}/2 - 2x + \ln(2)\]

  5. anonymous
    • 5 years ago
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    thanks guys is my first time here. I am lost in differentials. greetings from Barcelona

  6. anonymous
    • 5 years ago
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    No probs.

  7. anonymous
    • 5 years ago
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    I 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.

  8. anonymous
    • 5 years ago
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    \[y=(1/2)(x-4)(x)(y+1)+2\] The derivative of the right side of the above with respect to x is: \[(1/2)(x-4)(y+1)+(1/2) (x)(y+1)\] or \[x y+x-2 y-2\]

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