Question

1. given that ( w^{prime prime}(x)=5 e^{x}+2, w^{prime}(0)=3 ), and ( w(0)=9 ), find ( w(x) ).

Explanation:

Step1: Integrate \( w''(x) \) to find \( w'(x) \)

Integrate \( w''(x)=5e^{x}+2 \).
Using the integral rules \( \int e^{x}dx = e^{x}+C\) and \( \int kdx=kx + C\) (where \(k\) is a constant), we have:
\( w'(x)=\int(5e^{x}+2)dx=5e^{x}+2x + C_1\)

Step2: Use \( w'(0) = 3 \) to find \( C_1 \)

Substitute \( x = 0\) and \( w'(0)=3\) into \( w'(x)=5e^{x}+2x + C_1\).
\( 3=5e^{0}+2\times0 + C_1\)
Since \( e^{0}=1\), then \( 3 = 5 + C_1\), so \( C_1=- 2\).
Thus, \( w'(x)=5e^{x}+2x-2\)

Step3: Integrate \( w'(x) \) to find \( w(x) \)

Integrate \( w'(x)=5e^{x}+2x - 2\).
Using the integral rules \( \int e^{x}dx = e^{x}+C\), \( \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \( \int kdx=kx + C\) (where \(k\) is a constant), we get:
\( w(x)=\int(5e^{x}+2x - 2)dx=5e^{x}+x^{2}-2x + C_2\)

Step4: Use \( w(0)=9 \) to find \( C_2 \)

Substitute \( x = 0\) and \( w(0)=9\) into \( w(x)=5e^{x}+x^{2}-2x + C_2\).
\( 9=5e^{0}+0^{2}-2\times0 + C_2\)
Since \( e^{0}=1\), then \( 9 = 5+C_2\), so \( C_2 = 4\)

Answer:

\( w(x)=5e^{x}+x^{2}-2x + 4\)