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To find the coefficient of \( x^3 \) in the function \( h(x) = f(x+1) - g(x+2) \), we first need to expand each component.
Step 1: Expand \( f(x+1) \)
Starting with \( f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4 \), expand each term:
1. \( 10(x+1) = 10x + 10 \)
2. \( a_2(x+1)^2 = a_2(x^2 + 2x + 1) = a_2x^2 + 2a_2x + a_2 \)
3. \( a_3(x+1)^3 = a_3(x^3 + 3x^2 + 3x + 1) = a_3x^3 + 3a_3x^2 + 3a_3x + a_3 \)
4. \( (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 \)
Collecting terms for the \( x^3 \) coefficient in \( f(x+1) \):
\( a_3x^3 + 4x^3 = (a_3 + 4)x^3 \)
Step 2: Expand \( g(x+2) \)
Next, expand \( g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4 \):
1. \( 3(x+2) = 3x + 6 \)
2. \( b_2(x+2)^2 = b_2(x^2 + 4x + 4) = b_2x^2 + 4b_2x + 4b_2 \)
3. \( b_3(x+2)^3 = b_3(x^3 + 6x^2 + 12x + 8) = b_3x^3 + 6b_3x^2 + 12b_3x + 8b_3 \)
4. \( (x+2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16 \)
Collecting terms for the \( x^3 \) coefficient in \( g(x+2) \):
\( b_3x^3 + 8x^3 = (b_3 + 8)x^3 \)
Step 3: Calculate the coefficient of \( x^3 \) in \( h(x) \)
The coefficient of \( x^3 \) in \( h(x) = f(x+1) - g(x+2) \) is:
\((a_3 + 4) - (b_3 + 8) = a_3 + 4 - b_3 - 8 = a_3 - b_3 - 4\)
Step 4: Conclusion
Since \( f(x) \neq g(x) \) for any \( x \), it implies \( a_3 \neq b_3 \), the specific values don't make a difference to the meaning of this expression, the coefficient remains constant as:-4
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is