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anonymous

  • 5 years ago

how to solve dy/dt+2y=exp(t/2)

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  1. bahrom7893
    • 5 years ago
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    Use an inegrating factor: This equation is in the format: \[dy/dt + 2y = e^{t/2}\] \[dy/dt + P(x)y = Q(x)\]

  2. bahrom7893
    • 5 years ago
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    Use an inegrating factor: This equation is in the format: dy/dt + 2y = e^{t/2} dy/dt + P(t)y = Q(t) I.f. = e^(int[P(t)dt])

  3. bahrom7893
    • 5 years ago
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    I.f. = e^[Integral(2dt)] = e^(2t) Multiply your diff equation by the integrating factor e^(2t) e^(2t) * [dy/dt+2y] = e^(2t) * e^(t/2)

  4. bahrom7893
    • 5 years ago
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    I.f. = e^[Integral(2dt)] = e^(2t) Multiply your diff equation by the integrating factor e^(2t); e^(2t) * [dy/dt+2y] = e^(2t) * e^(t/2)

  5. bahrom7893
    • 5 years ago
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    Sorry for double post my pc is having issues. Now simplify: (dy/dt)*e^(2t) + 2y*e^(2t) = e^(2t + [t/2]); Notice the left side is a product rule for derivatives: (d/dt)[y*e^2t] = e^(5t/2); {Btw if you don't understand how i got the left side; differentiate that product with respect to t, you just have to recognize the form and the reason why we use integrating factors is because we want to put one side of the d.e. into that form: y*2*e^(2t) + e^(2t)*(dy/dt)}

  6. bahrom7893
    • 5 years ago
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    So now you know that: (d/dt)[y*e^(2t)]=e^(5t/2) {derivative of y*e^(2t) with respect to t is equal to e^(5t/2)} Integrate both sides with respect to t: (d/dt)[y*e^(2t)]=e^(5t/2) Int{ (d/dt)[y*e^(2t)] * dt}=Int{ e^(5t/2) * dt }

  7. bahrom7893
    • 5 years ago
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    Integral of the derivative is just the original function (i think it is called newton's method or something like that). So the integral of the left side is: y*e^(2t) = Integral(e^[5t/2]*dt); the right side is easy to integrate: Integral(e^[5t/2]*dt); Let u = 5t/2; then du = (5/2) dt; Multiply the integral by (2/5)*(5/2) - Basically im multiplying by one, i just need that 5/2 for the du; so: Integral(e^[5t/2]*dt) = (2/5)* Integral(e^u * du) = (2/5)*e^u +C

  8. bahrom7893
    • 5 years ago
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    We let u = 5t/2, so: y*e^(2t) = (2/5)*e^u +C y*e^(2t) = (2/5)*e^(5t/2) +C

  9. bahrom7893
    • 5 years ago
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    Simplify (divide everything by e^(2t)): y*e^(2t) = (2/5)*e^(5t/2) + C y = (2/5) * e^([5t/2] - [2t]) + C * e^(-2t) y = (2/5) * e^(t/2) + C*e^(-2t) <= Your answer

  10. bahrom7893
    • 5 years ago
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    Simplify (divide everything by e^(2t)): y*e^(2t) = (2/5)*e^(5t/2) + C; y = (2/5) * e^([5t/2] - [2t]) + C * e^(-2t); y = (2/5) * e^(t/2) + C*e^(-2t) <= Your answer

  11. bahrom7893
    • 5 years ago
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    Sorry for double post again, im havin issues with pc.. So that's the method u use to solve these types of problems. My answer should be correct, just check my arithmetic, i had a class early in the morning and have a break now solved this without a calculator, but i think there shouldn't be any mistakes, but just in case check the work. But that's how u solve those problems. Btw, my next class is differential equations!

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