The particular solution of dy/dx = −4xy², given that y(0) = 1, isdy/dx = −4xy², जबकि y(0) = 1, का विशेष हल है
dy/dx = −4xy², जबकि y(0) = 1, का विशेष हल है
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Correct answer: (A) (2x² + 1)y = 1(2x² + 1)y = 1
Explanation
Separate variables: dy/y² = −4x dx, so −1/y = −2x² + c. With y(0) = 1, c = −1, giving 1/y = 2x² + 1, i.e. (2x² + 1)y = 1.चरों को पृथक करें: dy/y² = −4x dx, इसलिए −1/y = −2x² + c। y(0) = 1 से c = −1, अतः 1/y = 2x² + 1, अर्थात् (2x² + 1)y = 1।