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Find out the differential equation of `y=Acosx^2+Bsinx^2` |
Answer» `y = Acosx^2+Bsinx^2` `=>dy/dx = A(-sinx^2)(2x)+B(cosx^2)(2x)` `=>dy/dx = 2x(Bcosx^2 - Asinx^2)` `=>(d^2y)/dx^2 = 2x(-Bsinx^2(2x) - Acosx^2(2x))+(Bcosx^2 - Asinx^2)(2)` `=>(d^2y)/dx^2 = -4x^2(Acosx^2+Bsinx^2) +2(Bcosx^2 - Asinx^2)` `=>(d^2y)/dx^2 = -4x^2y +1/x(dy/dx)` `=>(d^2y)/dx^2 -1/x(dy/dx) + 4x^2y = 0`, which is the required differential equation. |
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