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(x-2y) dy/dx + (2x+y) = 0 , y(1)=1 then find the value y(2) ?
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(x-2y) dy/dx + (2x+y) = 0 , y(1)=1 then find the value y(2) ?
Solution:
1. Given equation:
The given differential equation is (x-2y) dy/dx + (2x+y) = 0.
2. Initial Condition:
The initial condition is y(1) = 1.
3. First Step - Rewrite the Equation:
Rewrite the given equation in the standard form of a first-order linear differential equation: dy/dx + (2x+y)/(x-2y) = 0.
4. Multiplying by Integrating Factor:
The integrating factor is e^∫(2/(x-2y))dx = e^(2ln|x-2y|) = (x-2y)^2.
Multiply the entire equation by (x-2y)^2 to get: (x-2y)^2 dy/dx + 2(x-2y)(2x+y) = 0.
5. Integrate the Equation:
Integrate the above equation to get: (x-2y)^3 = C, where C is the constant of integration.
6. Using the Initial Condition:
Substitute the initial condition y(1) = 1 into the equation to find the value of C: (1-2(1))^3 = C => C = -1.
7. Find y(2):
Now, substitute x=2 into the equation: (2-2y)^3 = -1.
Solve the above equation to find the value of y when x=2.
Therefore, y(2) = 1/2.

Conclusion:
The value of y(2) is 1/2.
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(x-2y) dy/dx + (2x+y) = 0 , y(1)=1 then find the value y(2) ?
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