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2. Consider the first order linear equation \[ p(x) \frac{d y}{d x}-q(x) y=h(x)...
Sep 22, 2024
Solution by Steps
step 2
The integrating factor is typically of the form e∫q(x)p(x)dx e^{\int \frac{q(x)}{p(x)} dx} or e−∫q(x)p(x)dx e^{-\int \frac{q(x)}{p(x)} dx}
step 3
From the options provided, we can see that e−∫q(x)dx e^{-\int q(x) dx} is a common integrating factor for the standard form of the linear equation
step 4
Therefore, the appropriate integrating factor to use is e−∫q(x)dx e^{-\int q(x) dx}
B
Key Concept
Integrating Factor for First Order Linear Equations
Explanation
The integrating factor is used to simplify the process of solving first order linear differential equations, allowing us to express the solution in a more manageable form. The standard form typically leads to the integrating factor e−∫q(x)dx e^{-\int q(x) dx} .
给我详细的解答过程，每一步都要好好写出来
Solution by Steps
step 1
We start with the differential equation (1+x2)dydx−2xy=2x (1 + x^2) \frac{dy}{dx} - 2xy = 2x . To solve it, we can rearrange it into standard form: dydx−2x1+x2y=2x1+x2 \frac{dy}{dx} - \frac{2x}{1+x^2} y = \frac{2x}{1+x^2}
step 2
Next, we identify p(x)=−2x1+x2 p(x) = -\frac{2x}{1+x^2} and h(x)=2x1+x2 h(x) = \frac{2x}{1+x^2} . The integrating factor μ(x) \mu(x) is given by e∫p(x) dx e^{\int p(x) \, dx} . We compute ∫−2x1+x2 dx \int -\frac{2x}{1+x^2} \, dx
step 3
The integral ∫−2x1+x2 dx \int -\frac{2x}{1+x^2} \, dx simplifies to −ln⁡(1+x2)+C -\ln(1+x^2) + C . Thus, the integrating factor is μ(x)=e−ln⁡(1+x2)=11+x2 \mu(x) = e^{-\ln(1+x^2)} = \frac{1}{1+x^2}
step 4
We multiply the entire differential equation by the integrating factor: 11+x2((1+x2)dydx−2xy)=2x1+x2 \frac{1}{1+x^2} \left( (1+x^2) \frac{dy}{dx} - 2xy \right) = \frac{2x}{1+x^2} . This simplifies to dydx−2x1+x2y=2x(1+x2)2 \frac{dy}{dx} - \frac{2x}{1+x^2} y = \frac{2x}{(1+x^2)^2}
step 5
The left-hand side can be expressed as ddx(y1+x2) \frac{d}{dx} \left( \frac{y}{1+x^2} \right) . Therefore, we have ddx(y1+x2)=2x(1+x2)2 \frac{d}{dx} \left( \frac{y}{1+x^2} \right) = \frac{2x}{(1+x^2)^2}
step 6
Integrating both sides gives us y1+x2=∫2x(1+x2)2 dx \frac{y}{1+x^2} = \int \frac{2x}{(1+x^2)^2} \, dx . The integral evaluates to −11+x2+C -\frac{1}{1+x^2} + C . Thus, y1+x2=−11+x2+C \frac{y}{1+x^2} = -\frac{1}{1+x^2} + C
step 7
Multiplying through by 1+x2 1+x^2 yields y=−1+C(1+x2) y = -1 + C(1+x^2) . To find C C , we use the boundary condition y(0)=1 y(0) = 1 : 1=−1+C(1+02) 1 = -1 + C(1+0^2) , which gives C=2 C = 2
step 8
Therefore, the solution to the differential equation is y=−1+2(1+x2)=2+2x2 y = -1 + 2(1+x^2) = 2 + 2x^2
Answer
The solution to the differential equation is y=2+2x2 y = 2 + 2x^2
Key Concept
Solving first-order linear differential equations using integrating factors
Explanation
We rearranged the equation, found the integrating factor, and applied it to solve for y y using the given boundary condition.
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