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Exercise

dydx=(e2x−3y)m+2\frac{dy}{dx}=\frac{\left(e^{2x-3y}\right)}{m+2}

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Final answer to the exercise

y=ln⁡(3(e2x+C1)2)3y=\frac{\ln\left(\frac{3\left(e^{2x}+C_1\right)}{2}\right)}{3}

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dydx =(e2x−3y)m+2 
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