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Consider the Initial Value Problem

What Is the Solution y=203e32t143e34t y=\frac{20}{3} e^{-\frac{3}{2} t}-\frac{14}{3} e^{-\frac{3}{4} t}

Question 28

Multiple Choice

Consider the initial value problem
 Consider the initial value problem   What is the solution of this initial value problem? A)    y=\frac{20}{3} e^{-\frac{3}{2} t}-\frac{14}{3} e^{-\frac{3}{4} t}   B)    y=\frac{4}{3} e^{\frac{3}{2} t}-\frac{2}{3} e^{\frac{3}{4} t}   C)    y=-\frac{14}{3} e^{-\frac{3}{2} t}+\frac{20}{3} e^{-\frac{3}{4} t}   D)    y=-\frac{2}{3} e^{\frac{3}{2} t}+\frac{4}{3} e^{\frac{3}{4} t}
What is the solution of this initial value problem?


A) y=203e32t143e34t y=\frac{20}{3} e^{-\frac{3}{2} t}-\frac{14}{3} e^{-\frac{3}{4} t}
B) y=43e32t23e34t y=\frac{4}{3} e^{\frac{3}{2} t}-\frac{2}{3} e^{\frac{3}{4} t}
C) y=143e32t+203e34t y=-\frac{14}{3} e^{-\frac{3}{2} t}+\frac{20}{3} e^{-\frac{3}{4} t}
D) y=23e32t+43e34t y=-\frac{2}{3} e^{\frac{3}{2} t}+\frac{4}{3} e^{\frac{3}{4} t}

Correct Answer:

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