Simple Differential EquationsI want to solve the two following differential equations:(1) f ″ ( t ) = 3 f ′ ( t ) − f ( t )(2) f ″ ( t ) = 2 f ′ ( t ) − f ( t )I chose the approach f ( t ) = e λ t and hence arrive for the first case at λ 2 e λ t = e λ t ( 3 λ − 1 ) → λ 2 = 3 λ − 1and for the second differential equation I arrive at λ 2 = 2 λ − 1Is this correct? How do I find λ now per hand easily? Also, in the exercise it says "find the solution and determine what λ is" - but isn't that the same thing?Also: For the second case, is t ∗ e λ t also a solution?

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Simple Differential EquationsI want to solve the two following differential equations:(1)       f    ″    (  t  )  =  3      f    ′    (  t  )  −  f  (  t  )(2)       f    ″    (  t  )  =  2      f    ′    (  t  )  −  f  (  t  )I chose the approach   f  (  t  )  =      e          λ      t       and hence arrive for the first case at      λ    2        e          λ      t        =      e          λ      t        (  3  λ  −  1  )  →      λ    2    =  3  λ  −  1and for the second differential equation I arrive at       λ    2    =  2  λ  −  1Is this correct? How do I find   λ now per hand easily? Also, in the exercise it says &quot;find the solution and determine what   λ is&quot; - but isn&#039;t that the same thing?Also: For the second case, is   t  ∗      e          λ      t       also a solution?

Lily Travis · Accepted Answer

Step 1There are some ways to find   λ easily such as the quadratic formula or doing some algebra. In this case, let&#039;s do the last one, so, for example:      λ    2    −  2  λ  +  1  =  0    (  λ  −  1      )    2    =  0    λ  =  1But   λ is not the solution of the differential equation. Instead:  f  (  x  )  =      C    1        e                  λ        1            x        +      C    2        e                  λ        2            x        +  ⋯  +      C    n        e                  λ        n            x      Step 2Where       C    i   are constants. That&#039;s true if       λ    1    ≠      λ    2    ≠  ⋯  ≠      λ    n  . In our case,       λ    1    =      λ    2  , so both solutions would be linearly dependent. To fix that, the solution should be:  f  (  x  )  =      C    1        e                  λ        1            x        +  x        C    2        e                  λ        1            x        +  ⋯  +      x    n          C    n        e                  λ        1            x      Where n is the multiplicity of   λ

I want to solve the two following differential equations: (1) f′′(t)=3f′(t)-f(t). (2) f′′(t)=2f′(t)-f(t)

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