[tex]y(t) = a exp(2t) + b exp(5t)[/tex]
[tex]y(0) = 2
[/tex] and [tex]y'(0) = 3
[/tex]
Find a and b. I did the substitutions and everything and got [tex]a= \frac{5}{3}
[/tex] and [tex]b= \frac{-1}{3}
[/tex] but apparently this is not right

Respuesta :

[tex]y(0)=2\implies 2=ae^{2(0)}+be^{5(0)}\implies 2=a+b[/tex]

[tex]y'(t)=2ae^{2t}+5be^{5t}[/tex]
[tex]y'(0)=3\implies 3=2ae^{2(0)}+5be^{5(0)}\implies 3=2a+5b[/tex]

Solving this should yield [tex]a=\dfrac73[/tex] and [tex]b=-\dfrac13[/tex].
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