btech m2 unit-1 imp Questions: • btech m2 unit-1,2,3,4,...
Solve the differential equation (x^2 + y^2) dx - 2xydy = 0.
The given differential equation is: (x^3 - y^2 + 2x)dx + 2ydy = 0 To find: The integrating factor of the given differential equation. Solution: To find the integrating factor, we can use the following formula: Integrating factor (IF) = e^∫P(x)dx Step 1: Identify P(x) in the given differential equation: The given differential equation is: (x^3 - y^2 + 2x)dx + 2ydy = 0 Comparing this with the standard form: M(x, y)dx + N(x, y)dy = 0 We can identify P(x) as: P(x) = x^3 - y^2 + 2x Step 2: Find the integral of P(x): ∫P(x)dx = ∫(x^3 - y^2 + 2x)dx Using the power rule of integration, we can find the integral: ∫x^3dx = (1/4)x^4 + C1 ∫-y^2dx = -y^2x + C2 ∫2xdx = x^2 + C3 So, ∫P(x)dx = (1/4)x^4 - y^2x + x^2 + C Step 3: Find the integrating factor: The integrating factor (IF) is given by the formula: IF = e^∫P(x)dx Substituting the value of ∫P(x)dx, we get: IF = e^((1/4)x^4 - y^2x + x^2 + C) Since C is an arbitrary constant, we can write it as: IF = Ke^((1/4)x^4 - y^2x + x^2) Where K is an arbitrary constant. Thus, the integrating factor of the given differential equation is: IF = Ke^((1/4)x^4 - y^2x + x^2)
Негізгі бет btech m2 unit-1|non exact differential equ|methods of non exact|methods 3|(x^2+y^2+2x)dx+2ydy=0
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