I'm looking at the following problem. I'm trying to solve for y, and separating variables doesn't look like an option. I'm thinking putting it into standard form of a first order linear eq would be a good place to start. Is this correct? y ′ − e − t + 3 y = − 3 y ( 0 ) = 5In standard form: y ′ + P ( x ) y = Q ( x ) y ′ + 3 y = − 3 + e − t y e 3 t = ∫ − 3 e 3 t + e 2 t ∗ d t y e 3 t = e − 3 t + 0.5 e 2 t + C y = − 1 + 0.5 e − t + CSolving for initial condition y(0) = 5: y = − 1 + 0.5 e − t + C y ( 0 ) = − 1 + 0.5 e 0 + C = 5 y ( 0 ) = − 0.5 + C = 5 y ( 0 ) = C = 5.5Plug back in C: y = − 1 + 0.5 e − t + 5.5I've went through the steps of calculating the integrating factor and solving for y. The answer was incorrect. Is this the right way to approach this problem?

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I&#039;m looking at the following problem. I&#039;m trying to solve for y, and separating variables doesn&#039;t look like an option. I&#039;m thinking putting it into standard form of a first order linear eq would be a good place to start. Is this correct?                    y        ′            −              e                  −          t                    +      3        y    =  −  3  y  (  0  )  =  5In standard form:      y    ′    +  P  (  x  )  y  =  Q  (  x  )      y    ′    +  3  y  =  −  3  +      e          −      t        y      e          3      t        =  ∫      −    3          e              3        t              +          e              2        t              ∗    d    t    y      e          3      t        =      e          −      3      t        +  0.5      e          2      t        +  C  y  =  −  1  +  0.5      e          −      t        +  CSolving for initial condition y(0) = 5:  y  =  −  1  +  0.5      e          −      t        +  C  y  (  0  )  =  −  1  +  0.5      e          0        +  C  =  5  y  (  0  )  =  −  0.5  +  C  =  5  y  (  0  )  =  C  =  5.5Plug back in C:  y  =  −  1  +  0.5      e          −      t        +  5.5I&#039;ve went through the steps of calculating the integrating factor and solving for y. The answer was incorrect. Is this the right way to approach this problem?

Valeria Wolfe · Accepted Answer

y        ′            −              e                  −          t                    +      3        y    =  −  3  ⟹      y    ′    −      e          −      t        =  −  3  y  −  3  ⟹      y    ′    +  3  y  =  −  3  +      e          −      t      As you can see, when you have the first-order linear ODE,      y    ′    +  a  (  t  )  y  =  b  (  t  )So in this case, 3 = a(t). So       e          ∫      3      d      t        =      e          3      t            d          d      t        (      e          3      t        y  )  =  −  3  +      e          −      t      Now integrate both sides AND DO NOT FORGET THE CONSTANT.      e          3      t        y  +  C  =  −  3  t  −      e          −      t      Isolate y. The end.

I'm looking at the following problem. I'm trying to solve for y, and separating variables doesn't lo

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