1. **Problem statement:** Solve the first order differential equation $$\frac{dy}{dt} + \frac{2y}{t+10} = 3$$ using an integrating factor and show the general solution is $$y = t + 10 + \frac{c}{(t+10)^2}$$ where $c$ is an arbitrary constant. 2. **Identify the integrating factor:** The equation is linear of the form $$\frac{dy}{dt} + P(t)y = Q(t)$$ with $$P(t) = \frac{2}{t+10}$$ and $$Q(t) = 3$$. 3. The integrating factor $$\mu(t)$$ is given by $$\mu(t) = e^{\int P(t) dt} = e^{\int \frac{2}{t+10} dt} = e^{2 \ln|t+10|} = (t+10)^2$$. 4. Multiply the entire differential equation by the integrating factor: $$ (t+10)^2 \frac{dy}{dt} + (t+10)^2 \frac{2y}{t+10} = 3 (t+10)^2 $$ Simplify the second term: $$ (t+10)^2 \frac{2y}{t+10} = 2y (t+10) $$ So the equation becomes: $$ (t+10)^2 \frac{dy}{dt} + 2y (t+10) = 3 (t+10)^2 $$ 5. Recognize the left side as the derivative of a product: $$ \frac{d}{dt} \left[ y (t+10)^2 \right] = 3 (t+10)^2 $$ 6. Integrate both sides with respect to $t$: $$ y (t+10)^2 = \int 3 (t+10)^2 dt + c $$ Calculate the integral: $$ \int 3 (t+10)^2 dt = 3 \int (t+10)^2 dt = 3 \cdot \frac{(t+10)^3}{3} = (t+10)^3 $$ So: $$ y (t+10)^2 = (t+10)^3 + c $$ 7. Divide both sides by $(t+10)^2$: $$ y = \frac{(t+10)^3 + c}{(t+10)^2} = t + 10 + \frac{c}{(t+10)^2} $$ This matches the required general solution. --- 1. **Problem statement:** Find the general solution of the second order differential equation: $$ \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} - 4y = 8x^2 $$ 2. **Solve the homogeneous equation:** $$ \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} - 4y = 0 $$ Characteristic equation: $$ r^2 - 3r - 4 = 0 $$ 3. Solve the quadratic: $$ r = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} $$ So roots are: $$ r_1 = 4, \quad r_2 = -1 $$ 4. Homogeneous solution: $$ y_h = c_1 e^{4x} + c_2 e^{-x} $$ 5. **Find a particular solution $y_p$:** Since the right side is a polynomial $8x^2$, try a polynomial of degree 2: $$ y_p = Ax^2 + Bx + C $$ 6. Compute derivatives: $$ y_p' = 2Ax + B $$ $$ y_p'' = 2A $$ 7. Substitute into the differential equation: $$ 2A - 3(2Ax + B) - 4(Ax^2 + Bx + C) = 8x^2 $$ Expand: $$ 2A - 6Ax - 3B - 4Ax^2 - 4Bx - 4C = 8x^2 $$ Group terms by powers of $x$: $$ (-4A) x^2 + (-6A - 4B) x + (2A - 3B - 4C) = 8x^2 + 0x + 0 $$ 8. Equate coefficients: - For $x^2$: $$ -4A = 8 \implies A = -2 $$ - For $x$: $$ -6A - 4B = 0 \implies -6(-2) - 4B = 0 \implies 12 - 4B = 0 \implies B = 3 $$ - For constant term: $$ 2A - 3B - 4C = 0 \implies 2(-2) - 3(3) - 4C = 0 \implies -4 - 9 - 4C = 0 \implies -13 - 4C = 0 \implies C = -\frac{13}{4} $$ 9. Particular solution: $$ y_p = -2x^2 + 3x - \frac{13}{4} $$ 10. **General solution:** $$ y = y_h + y_p = c_1 e^{4x} + c_2 e^{-x} - 2x^2 + 3x - \frac{13}{4} $$