10. 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\neq - 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)