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Solve the IVP (2x − 5y) dx + (4x − y) dy = 0, y(1) = 4.

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To solve the initial value problem (IVP) (2x − 5y) dx + (4x − y) dy = 0 with the initial condition y(1) = 4, we'll use the method of separation of variables.

Step 1: Separate the variables x and y on opposite sides of the equation:
(2x - 5y) dx = (y - 4x) dy

Step 2: Integrate both sides:
∫(2x - 5y) dx = ∫(y - 4x) dy

Step 3: Evaluate the integrals:
∫2x dx - ∫5y dx = ∫y dy - ∫4x dy

Step 4: Continue solving the integrals:
x^2 - 5xy = y^2 - 4xy + C

Step 5: Rearrange the equation and simplify:
x^2 - 5xy + 4xy - y^2 = C

Step 6: Combine like terms:
x^2 - y^2 - xy = C

Step 7: Use the initial condition y(1) = 4 to find the value of C:
(1)^2 - (4)^2 - 1(4) = C
1 - 16 - 4 = C
C = -19

Step 8: Write the final equation:
x^2 - y^2 - xy = -19

This is the solution to the initial value problem (IVP) (2x − 5y) dx + (4x − y) dy = 0 with the initial condition y(1) = 4.
User Rick Regan
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