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guess Group Title

dy/dx+y=x^2+2x solve it by 2 method

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    \[\large y'+y=x^2+2x\] Let's start by using the method of Undetermined Coefficients. Our complimentary solution \(\large y_c\) is given by the solution to this equation.\[\large y'+y=0\] The characteristic equation is,\[\large r+1=0 \qquad \rightarrow \qquad r=-1\]Giving us a complimentary solution of, \(\large \qquad y_c=Ce^{-x}\) ---------------------------------------------------------- We will also need to find the particular solution \(\large y_p\). Our particular solution will be of the same form as the right side of our problem. In this case, we have a polynomial of degree 2. So our particular solution will look something like,\[\large y_p=Ax^2+B(2x)+C\]Where A, B and C are unknown constants. We'll take the derivative of our particular solution and then we can plug in \(\large y_p\) and \(\large y_p^{'}\) into the original problem to solve for A, B and C.

    • one year ago
  2. zepdrix Group Title
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    Eventually we're trying to get an answer that consists of the complimentary AND particular solution. \(\large \qquad y=y_c+y_p\)

    • one year ago
  3. zepdrix Group Title
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    \[\large y_p=Ax^2+2Bx+C\]\[\large y_p^{'}=2Ax+2B\]

    • one year ago
  4. zepdrix Group Title
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    Plugging these into the original equation gives us,\[y'+y=x^2+2x \qquad \rightarrow \qquad \left(2Ax+2B\right)+\left(Ax^2+2Bx+C\right)=x^2+2x\]

    • one year ago
  5. zepdrix Group Title
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    From here we can equate like terms.\[2B+C=0\]\[2Ax+2Bx=2x \qquad \rightarrow \qquad 2A+2B=2\]\[Ax^2=x^2 \qquad \rightarrow \qquad A=1\]

    • one year ago
  6. zepdrix Group Title
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    Does any of this make sense @guess or should i stop? XD

    • one year ago
  7. guess Group Title
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    @zepdrix thank you ,but if you can solve it by laplace Be very grateful

    • one year ago
  8. guess Group Title
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    @zepdrix yes the particular solution i think that what he want thanks and don't stop please:)

    • one year ago
  9. nitz Group Title
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    u need any help........

    • one year ago
  10. guess Group Title
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    @nitz yes please

    • one year ago
  11. hartnn Group Title
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    u know whats L[dy/dx] =... ?

    • one year ago
  12. hartnn Group Title
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    assuming initial conditions =0 L[dy/dx] = s Y(s) where Y(s) is the Laplace Transform of y(t). so, take Laplace on both sides of dy/dx+y=x^2+2x and can you tell me what you get in first step ??

    • one year ago
  13. guess Group Title
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    @zepdrix then what ? is b=1/2 ?

    • one year ago
  14. guess Group Title
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    @zepdrix @hartnn @nitz this question came at midterm and ididn't solve it ): help me please

    • one year ago
  15. Callisto Group Title
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    Can't we use integrating factor to solve it?

    • one year ago
  16. guess Group Title
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    @Callisto mm Can you show me?

    • one year ago
  17. mukushla Group Title
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    what Callisto said probabely is the best bet :)

    • one year ago
  18. Callisto Group Title
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    \[y'+p(x)y=q(x)\] \[\alpha = exp[\int p(x)dx]\]\[y=\frac{1}{\alpha}\int \alpha [q(x)] dx\]

    • one year ago
  19. Callisto Group Title
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    in your case p(x)=1, q(x) = x^2+2x

    • one year ago
  20. guess Group Title
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    @Callisto it look like will be best bet really !! continue please :)

    • one year ago
  21. Callisto Group Title
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    Ha! Several integration by parts there :S \[y'+y=x^2+2x\] \[\alpha = e^{\int 1 dx}=e^x\] \[y=e^{-x}\int e^x(x^2+2x)dx\]\[=e^{-x}[e^x(x^2+2x)-\int e^x(2x+2)dx]\]\[=e^{-x}[e^x(x^2+2x)-2(e^x(x+1)-\int e^xdx)]\]\[=...\]

    • one year ago
  22. Callisto Group Title
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    @guess Can you do it from here? Or still need more help?

    • one year ago
  23. hartnn Group Title
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    using Laplace \(L[dy/dx]+L[y]=L[x^2]+2L[x] \\sY+Y=2/s^3+2/s \\ Y=2(1+s^2)/[s^3(s+1)] \\ Y=2/x^3-2/x^2+4/x-4/(s+1)\\\text{skipping the partial fractions part for you.Taking Inverse Laplace}\\ y = x^2-2x+4-4e^x\) Someone please verify the final answer @zepdrix or @Callisto

    • one year ago
  24. hartnn Group Title
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    \(y=e^{-x}\int e^x(x^2+2x)dx , after \:\: simplification, y=x^2\)

    • one year ago
  25. guess Group Title
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    @hartnn @Callisto yes please continue i got this put i feel it Error exp^(- x) {x^2e^x-2e^x} and thank alot for all

    • one year ago
  26. Callisto Group Title
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    Method of integrating factor: y'+y=x^2+2x \[\alpha =e^x\]\[y=e^{-x}\int e^x(x^2+2x)dx\]\[=e^{-x}[e^x(x^2+2x)-2\int e^x(x+1)]\]\[=e^{-x}[e^x(x^2+2x)-2e^x(x+1)+2e^x+C]\]\[=e^{-x}(e^xx^2+C)\]\[=x^2+Ce^{-x}\]

    • one year ago
  27. guess Group Title
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    thank you so much @Callisto @hartnn you're very helper :))

    • one year ago
  28. guess Group Title
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    by laplace L[ y`]+L[y]=L[x2]+2L[x] S(Y)+Y(S)=2/S^3+2/S^2 (S+1)Y(S)=2(S+1)/S^3 y(S)=2/S^3 y(x)=x^2 @hartnn @Callisto right ??

    • one year ago
  29. hartnn Group Title
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    ohh.....yes!

    • one year ago
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