Math  /  Calculus

QuestionFind the particular solution of the differential equation that satisfies the initial condition(s). f(x)=ex,f(0)=5,f(0)=9f(x)=\begin{array}{l} f^{\prime \prime}(x)=e^{x}, \quad f^{\prime}(0)=5, \quad f(0)=9 \\ f(x)=\square \end{array} Need Help? Read It Watch It Submit Answer

Studdy Solution

STEP 1

1. We are given a second-order differential equation f(x)=ex f''(x) = e^x .
2. We have initial conditions f(0)=5 f'(0) = 5 and f(0)=9 f(0) = 9 .
3. We need to find the particular solution f(x) f(x) .

STEP 2

1. Integrate the second derivative to find the first derivative.
2. Use the initial condition for the first derivative to find the constant of integration.
3. Integrate the first derivative to find the function f(x) f(x) .
4. Use the initial condition for the function to find the second constant of integration.
5. Write the particular solution.

STEP 3

Integrate the second derivative f(x)=ex f''(x) = e^x with respect to x x to find the first derivative f(x) f'(x) :
f(x)=exdx=ex+C1 f'(x) = \int e^x \, dx = e^x + C_1

STEP 4

Use the initial condition f(0)=5 f'(0) = 5 to solve for C1 C_1 :
f(0)=e0+C1=5 f'(0) = e^0 + C_1 = 5 1+C1=5 1 + C_1 = 5 C1=4 C_1 = 4

STEP 5

Now integrate the first derivative f(x)=ex+4 f'(x) = e^x + 4 to find the function f(x) f(x) :
f(x)=(ex+4)dx=ex+4x+C2 f(x) = \int (e^x + 4) \, dx = e^x + 4x + C_2

STEP 6

Use the initial condition f(0)=9 f(0) = 9 to solve for C2 C_2 :
f(0)=e0+40+C2=9 f(0) = e^0 + 4 \cdot 0 + C_2 = 9 1+C2=9 1 + C_2 = 9 C2=8 C_2 = 8

STEP 7

Write the particular solution for f(x) f(x) :
f(x)=ex+4x+8 f(x) = e^x + 4x + 8
The particular solution is:
f(x)=ex+4x+8 \boxed{f(x) = e^x + 4x + 8}

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