find the general and particular solution of the differential equation 
d^2 y/dx^2 =8x+10 , f'(1)=3, f(2)=5

To find the general and particular solutions of the differential equation 

\[
\frac{d^2 y}{dx^2} = 8x + 10
\]

with the initial conditions \( f'(1) = 3 \) and \( f(2) = 5 \), we proceed step-by-step.

### Step 1: Find the General Solution

1. **Integrate the differential equation**: 

   The given equation is a second-order differential equation. To solve it, we first find the first derivative by integrating the right-hand side with respect to \(x\):

   \[
   \frac{dy}{dx} = \int (8x + 10) \, dx
   \]

2. **Compute the first integration**:

   \[
   \frac{dy}{dx} = 8 \int x \, dx + 10 \int dx
   \]

   \[
   \frac{dy}{dx} = 8 \left(\frac{x^2}{2}\right) + 10x + C_1
   \]

   \[
   \frac{dy}{dx} = 4x^2 + 10x + C_1
   \]

   Here, \(C_1\) is the constant of integration.

3. **Integrate again to find \(y(x)\)**:

   \[
   y(x) = \int (4x^2 + 10x + C_1) \, dx
   \]

   \[
   y(x) = 4 \int x^2 \, dx + 10 \int x \, dx + C_1 \int dx
   \]

   \[
   y(x) = 4 \left(\frac{x^3}{3}\right) + 10 \left(\frac{x^2}{2}\right) + C_1x + C_2
   \]

   \[
   y(x) = \frac{4x^3}{3} + 5x^2 + C_1x + C_2
   \]

   Here, \(C_2\) is another constant of integration.

The **general solution** is:

\[
y(x) = \frac{4x^3}{3} + 5x^2 + C_1x + C_2
\]

### Step 2: Find the Particular Solution

Now we use the initial conditions \( f'(1) = 3 \) and \( f(2) = 5 \) to find \(C_1\) and \(C_2\).

1. **Apply the first initial condition \( f'(1) = 3 \)**:

   We know \( \frac{dy}{dx} = 4x^2 + 10x + C_1 \). So, plugging in \(x = 1\):

   \[
   4(1)^2 + 10(1) + C_1 = 3
   \]

   \[
   4 + 10 + C_1 = 3
   \]

   \[
   14 + C_1 = 3 \implies C_1 = 3 - 14 = -11
   \]

2. **Apply the second initial condition \( f(2) = 5 \)**:

   Substituting \(x = 2\) into the general solution:

   \[
   y(2) = \frac{4(2)^3}{3} + 5(2)^2 - 11(2) + C_2
   \]

   \[
   y(2) = \frac{4 \cdot 8}{3} + 5 \cdot 4 - 22 + C_2
   \]

   \[
   y(2) = \frac{32}{3} + 20 - 22 + C_2
   \]

   \[
   y(2) = \frac{32}{3} - 2 + C_2
   \]

   Set \(y(2) = 5\):

   \[
   5 = \frac{32}{3} - 2 + C_2
   \]

   To simplify, convert \(-2\) to a fraction:

   \[
   5 = \frac{32}{3} - \frac{6}{3} + C_2
   \]

   \[
   5 = \frac{26}{3} + C_2
   \]

   Subtract \(\frac{26}{3}\) from both sides:

   \[
   C_2 = 5 - \frac{26}{3}
   \]

   Convert 5 to a fraction:

   \[
   C_2 = \frac{15}{3} - \frac{26}{3} = -\frac{11}{3}
   \]

Thus, \(C_2 = -\frac{11}{3}\).

### Conclusion: The Particular Solution

Substitute \(C_1 = -11\) and \(C_2 = -\frac{11}{3}\) back into the general solution:

\[
y(x) = \frac{4x^3}{3} + 5x^2 - 11x - \frac{11}{3}
\]

This is the **particular solution** of the differential equation with the given initial conditions.

---

Do you want more details, or do you have any questions?

Here are some related questions you might find interesting:

1. How do we solve a second-order differential equation with constant coefficients?
2. What are the methods for solving non-homogeneous differential equations?
3. How does the choice of initial conditions affect the particular solution of a differential equation?
4. How can we verify if a solution to a differential equation is correct?
5. What are the applications of differential equations in physics and engineering?

**Tip:** Always check your solutions by substituting them back into the original differential equation and verifying the initial conditions.

Explore the general and particular solutions of the differential equation d^2 y/dx^2 = 8x + 10, with detailed steps and initial conditions f'(1) = 3 and f(2) = 5. This problem covers second-order differential equations and initial value problems, suitable for college-level students.

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