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dy/dx=y+(integration 0 to 1 ydx) given Y=1, where x=0
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dy/dx=y+(integration 0 to 1 ydx) given Y=1, where x=0
Understanding the Differential Equation
The equation given is dy/dx = y + ∫(from 0 to 1) y dx. This is a first-order linear ordinary differential equation that involves an integral term.
Initial Condition
We have the initial condition Y(0) = 1. This means when x = 0, the value of y is 1.
Step 1: Solve the Integral
- The integral ∫(from 0 to 1) y dx is a constant with respect to x, since y does not depend on x in this context.
- Let's denote the integral as C = ∫(from 0 to 1) y dx.
Step 2: Substitute the Integral into the Equation
- The differential equation now becomes dy/dx = y + C.
- This is a linear first-order differential equation which can be solved using the integrating factor method.
Step 3: Find the Integrating Factor
- The integrating factor, μ(x), is given by e^(∫(1 dx)) = e^x.
Step 4: Solve the Equation
- Multiplying the entire equation by the integrating factor gives:
e^x * dy/dx - e^x * y = C * e^x.
- Integrating both sides will yield the solution for y in terms of x.
Step 5: Apply Initial Condition
- Substitute x = 0 into the solution to find C using the initial condition Y(0) = 1.
Conclusion
By solving the differential equation step by step, we can find the specific function y(x) that satisfies both the equation and the initial condition. The key strategy is transforming the integral into a constant and then using integrating factors to simplify the equation.
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dy/dx=y+(integration 0 to 1 ydx) given Y=1, where x=0
is the final answer 2? maybe..
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dy/dx=y+(integration 0 to 1 ydx) given Y=1, where x=0
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