Solve the differential equation with constant coefficients

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Fatima Hasan
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Homework Statement



Printed text reading 'Solve the DE:' above the equation 2y'' - 2y' + 7y = 0.


Homework Equations



Typeset equation: y(x) = C1 exp(alpha x) cos(beta x) + C2 exp(alpha x) sin(beta x).

Handwritten or typeset expression lambda = alpha plus or minus beta i, showing a complex characteristic root of a differential equation.


The Attempt at a Solution


[/B]
Handwritten-style typeset working for a second-order linear ODE: 2y'' - 2y' + 7y = 0, auxiliary equation 2λ² - 2λ + 7 = 0, roots λ = (2 ± √(4-56))/4 = 1/2 ± √(56)i, and general solution y(x) = C1 exp(x/2)cos(√(56)x) + C2 exp(x/2)sin(√(56)x).

Is my answer correct?
 
Attachments
  • Printed text reading 'Solve the DE:' above the equation 2y'' - 2y' + 7y = 0.
    Capture.png
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  • Typeset equation: y(x) = C1 exp(alpha x) cos(beta x) + C2 exp(alpha x) sin(beta x).
    gif.gif
    1.3 KB · Views: 521
  • Handwritten or typeset expression lambda = alpha plus or minus beta i, showing a complex characteristic root of a differential equation.
    gif.gif
    340 bytes · Views: 556
  • Handwritten-style typeset working for a second-order linear ODE: 2y'' - 2y' + 7y = 0, auxiliary equation 2λ² - 2λ + 7 = 0, roots λ = (2 ± √(4-56))/4 = 1/2 ± √(56)i, and general solution y(x) = C1 exp(x/2)cos(√(56)x) + C2 exp(x/2)sin(√(56)x).
    gif.gif
    3.7 KB · Views: 567
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How do you get √56? Shouldn't it be (√52)/4 = (√13)/2?
 
mjc123 said:
How do you get √56? Shouldn't it be (√52)/4 = (√13)/2?
Worked solution of a second-order linear ODE: characteristic equation 2λ² - 2λ + 7 = 0, roots λ = 1/2 ± i√13/2, and general solution y(x) = C1 exp(x/2) cos(√13x/2) + C2 exp(x/2) sin(√13x/2).

Is it correct now ?
 
Attachments
  • Worked solution of a second-order linear ODE: characteristic equation 2λ² - 2λ + 7 = 0, roots λ = 1/2 ± i√13/2, and general solution y(x) = C1 exp(x/2) cos(√13x/2) + C2 exp(x/2) sin(√13x/2).
    gif.gif
    3.8 KB · Views: 569
Fatima Hasan said:
View attachment 234491
Is it correct now ?
1. In the future, please post questions about differential equations in the Calculus & Beyond section. I will move the ones you have posted in the Precalc section.
2. You don't need to ask us to check your work for these kinds of problems. Just substitute your solution in the original diff. equation.