Q. The particular solution of the differential equation dxdy=x2−xy+1, when x=2 and y=1 isA) xy=4x−6B) xy=2x−2C) xy=x−2D) xy=−x+4
Given Differential Equation: Given the differential equation dxdy=x2−xy+1 and the initial condition x=2, y=1. We need to find the particular solution that fits this condition.
Simplify Equation: Simplify the differential equation by factoring the denominator: dxdy=(x−1)xy+1.
Apply Initial Condition: Use the initial condition to plug in x=2 and y=1 into the equation: ((2−1)2)(1+1)=22=1. This confirms that the slope at x=2, y=1 is 1.
Integrate Differential Equation: To find the particular solution, we need to integrate the differential equation. We can separate variables and integrate: ∫y+1dy=∫x(x−1)dx.
Exponentiate to Solve for y: Integrate both sides: ln∣y+1∣=ln∣x∣−ln∣x−1∣+C, where C is the integration constant.
Find Integration Constant: Exponentiate both sides to solve for y: y+1=C⋅x/(x−1).
Substitute for Particular Solution: Use the initial condition (x=2,y=1) to find C: 1+1=C⋅(2−1)2, so C=1.
Finalize Solution: Substitute C back into the equation: y+1=x−1x. Solve for y: y=x−1x−1=x−1x−x+1=x−11.
Rewrite in Terms of x: Rewrite y in terms of x to match the answer choices: y=x−11. Multiply both sides by x: xy=x−1x.
Simplify the Expression: Simplify the expression: xy=x−1x=x−1x⋅x−1x−1=x−1x. This simplifies to xy=x−1.