1. **State the problem:** We want to solve the differential equation $$y' = ky$$ with $$k=1$$ using a power series approach. We assume the solution has the form $$y = f(x) = \sum_{i=0}^\infty a_i x^i$$ where $$a_i$$ are coefficients to be determined. 2. **Find the derivative of the power series:** Differentiating term-by-term, $$ f'(x) = \frac{d}{dx} \sum_{i=0}^\infty a_i x^i = \sum_{i=0}^\infty a_i \frac{d}{dx} x^i = \sum_{i=0}^\infty a_i i x^{i-1} = \sum_{i=1}^\infty i a_i x^{i-1} $$ Note the sum starts at $$i=1$$ because the $$i=0$$ term is zero. 3. **Substitute into the differential equation:** Since $$k=1$$, $$ f'(x) = f(x) \implies \sum_{i=1}^\infty i a_i x^{i-1} = \sum_{i=0}^\infty a_i x^i $$ 4. **Rewrite the right side to match powers:** Change index on right side to powers of $$x^{i-1}$$ by substituting $$j = i-1$$, $$ \sum_{i=0}^\infty a_i x^i = a_0 + \sum_{i=1}^\infty a_i x^i = a_0 + \sum_{i=1}^\infty a_i x^i $$ We want to compare coefficients of $$x^{i-1}$$, so rewrite right side as $$ \sum_{i=1}^\infty a_{i-1} x^{i-1} + a_0 x^{-1} $$ Since $$x^{-1}$$ term is not in left side, for equality, $$a_0 x^{-1}$$ must be zero, so ignore it for $$i \geq 1$$. 5. **Equate coefficients of like powers:** For each $$i \geq 1$$, $$ i a_i = a_{i-1} $$ 6. **Find coefficients recursively:** - For $$i=1$$: $$1 \cdot a_1 = a_0 \implies a_1 = a_0$$ - For $$i=2$$: $$2 a_2 = a_1 \implies a_2 = \frac{a_1}{2} = \frac{a_0}{2}$$ - For $$i=3$$: $$3 a_3 = a_2 \implies a_3 = \frac{a_2}{3} = \frac{a_0}{3!}$$ - For $$i=4$$: $$4 a_4 = a_3 \implies a_4 = \frac{a_3}{4} = \frac{a_0}{4!}$$ 7. **General formula:** By induction, $$ a_k = \frac{a_0}{k!} $$ 8. **Calculate $$a_{101}$$:** $$ a_{101} = \frac{a_0}{101!} $$ 9. **Write the series expansion for $$y$$:** $$ y = \sum_{i=0}^\infty a_i x^i = a_0 \sum_{i=0}^\infty \frac{x^i}{i!} = a_0 e^x $$ **Final answer:** $$ y = a_0 e^x $$ This matches the known solution to $$y' = y$$, confirming our approach.