Question
Show that $y=\frac{2}{3} e^x+e^{-2 x}$ is a solution of the differential equation $y^{\prime}+2 y=2 e^x$.
Step 1
The function is \( y = \frac{2}{3} e^x + e^{-2x} \). The differential equation is \( y' + 2y = 2e^x \). Show more…
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Show that $y=\frac{2}{3} e^{x}+e^{-2 x}$ is a solution of the differential equation $y^{\prime}+2 y=2 e^{x}$
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Show that $ y = \frac{2}{3}e^x + e^{-2x} $ is a solution of the differential equation $ y^x + 2y = 2e^x $.
Determine whether the given function is a solution of the differential equation. $$y=\frac{2}{3} e^{x}+e^{-2 x} ; \quad y^{\prime}+2 y=2 e^{x}$$
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