Find solution of y′(t)=ky(t) with y(1)=5y′(1)=4. y=Cekt
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Write as dydt=ky.
Therefore k=ln(y)−ct. When y=5,t=1,ln(5)=k+c.k=ln(5)−c.
When dydt=4,t=1:4=5(ln(5)−c).ln(5)−c=0.8, so c=ln(5)−0.8.
So y=e0.8t+ln(5)−0.8=5e0.8t(0.4493)=2.2466e0.8t approx.
See answers (2)
See answers (1)
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