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EXERCISE 1.5 · Q204

Q.eax+be^{ax+b}

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Let y=eax+by=e^{ax+b}.

Step 1: y1=aeax+by_1=ae^{ax+b}.

Step 2: y2=a2eax+by_2=a^2e^{ax+b}.

Step 3: y3=a3eax+by_3=a^3e^{ax+b}.

Pattern: every differentiation, by the chain rule, multiplies by one more factor of aa while the exponential …

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