Solve differential equation (x+1) dy/dx – 2y = (x+1)4
The Correct Answer and Explanation is:
We are given the differential equation: (x+1)dydx−2y=(x+1)4(x+1) \frac{dy}{dx} – 2y = (x+1)^4
Step 1: Rearranging into Standard Linear Form
We want to rewrite the equation in the form: dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)
Divide the entire equation by x+1x+1 (assuming x≠−1x \neq -1): dydx−2x+1y=(x+1)3\frac{dy}{dx} – \frac{2}{x+1}y = (x+1)^3
This is now a first-order linear differential equation with:
- P(x)=−2x+1P(x) = -\frac{2}{x+1}
- Q(x)=(x+1)3Q(x) = (x+1)^3
Step 2: Finding the Integrating Factor
The integrating factor (IF) is: μ(x)=e∫P(x)dx=e∫−2x+1dx=e−2ln∣x+1∣=1(x+1)2\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{2}{x+1} dx} = e^{-2 \ln|x+1|} = \frac{1}{(x+1)^2}
Step 3: Multiply Through by the Integrating Factor
Multiply both sides of the standard form by 1(x+1)2\frac{1}{(x+1)^2}: 1(x+1)2dydx−2(x+1)3y=(x+1)\frac{1}{(x+1)^2} \frac{dy}{dx} – \frac{2}{(x+1)^3}y = (x+1)
Left-hand side becomes the derivative of y⋅1(x+1)2y \cdot \frac{1}{(x+1)^2}: ddx(y(x+1)2)=(x+1)\frac{d}{dx}\left( \frac{y}{(x+1)^2} \right) = (x+1)
Step 4: Integrate Both Sides
ddx(y(x+1)2)=(x+1)\frac{d}{dx}\left( \frac{y}{(x+1)^2} \right) = (x+1)
Integrate both sides: y(x+1)2=∫(x+1)dx=(x+1)22+C\frac{y}{(x+1)^2} = \int (x+1) dx = \frac{(x+1)^2}{2} + C
Step 5: Solve for yy
y=(x+1)2((x+1)22+C)y = (x+1)^2 \left( \frac{(x+1)^2}{2} + C \right) y=(x+1)42+C(x+1)2\boxed{y = \frac{(x+1)^4}{2} + C(x+1)^2}
Explanation
The given differential equation (x+1)dydx−2y=(x+1)4(x+1)\frac{dy}{dx} – 2y = (x+1)^4 is a first-order linear differential equation. These types of equations are important in modeling systems where the rate of change of a quantity depends linearly on the quantity itself and possibly the independent variable—in this case, xx.
The first step involves recognizing the standard linear form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x). To get it into this form, we divide the entire equation by (x+1)(x+1), simplifying it while noting that x≠−1x \neq -1 (as division by zero is undefined).
Once in standard form, we find the integrating factor (IF). This factor transforms the left-hand side of the equation into the derivative of a product of functions, which is easier to integrate. The integrating factor here is μ(x)=1(x+1)2\mu(x) = \frac{1}{(x+1)^2}, derived from the exponential of the integral of P(x)P(x).
Multiplying through by the IF simplifies the equation, enabling us to apply the product rule in reverse: we recognize the left-hand side as a derivative. Integrating both sides then yields an expression for yy, including a constant of integration CC, which accounts for all possible solutions depending on initial conditions.
The final solution expresses yy as a function of xx, combining a particular solution to the nonhomogeneous part and the general solution of the associated homogeneous equation. This approach highlights how powerful and systematic linear differential equations and integrating factors are in solving real-world mathematical models.
