Solve it in video format ---b) The given solution of differential equation is y = eˣ(A cos x + B sin x) ....(i) Differentiating, we get dy/dx = eˣ(A cos x + B sin x) + eˣ(-A sin x + B cos x) = y + eˣ (-A sin x + B cos x) d²y/dx² = dy/dx + eˣ (-A sin x + B cos x) + eˣ (-A cos x - B sin x) = dy/dx + (dy/dx - y) - y = 2 dy/dx - 2y or, d²y/dx² - 2 dy/dx + 2y = 0, which is the required differential equation.

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We are given a solution to a differential equation in the form y equals e to the x times A cosine x plus B sine x. Our task is to find the differential equation that has this as its general solution. We will do this by differentiating the given solution twice and eliminating the arbitrary constants A and B. Now we begin the first step of finding the differential equation. We differentiate the given solution y equals e to the x times A cosine x plus B sine x with respect to x. Using the product rule, we get dy dx equals e to the x times A cosine x plus B sine x plus e to the x times negative A sine x plus B cosine x. We can simplify this by substituting the original y back into the expression, giving us dy dx equals y plus e to the x times negative A sine x plus B cosine x. Now we perform the second differentiation. We differentiate dy dx with respect to x to get d squared y dx squared. This equals d dx of y plus e to the x times negative A sine x plus B cosine x. Expanding this using the product rule, we get dy dx plus e to the x times negative A sine x plus B cosine x plus e to the x times negative A cosine x minus B sine x. Notice that the term e to the x times negative A sine x plus B cosine x equals dy dx minus y from our previous step. Also, the term e to the x times negative A cosine x minus B sine x equals negative y. Substituting these relationships, we get d squared y dx squared equals dy dx plus dy dx minus y minus y, which simplifies to d squared y dx squared equals 2 dy dx minus 2 y. Finally, we rearrange our result to get the standard form of the differential equation. From d squared y dx squared equals 2 dy dx minus 2 y, we move all terms to one side to get d squared y dx squared minus 2 dy dx plus 2 y equals zero. This is our required differential equation! We can verify that our original solution y equals e to the x times A cosine x plus B sine x indeed satisfies this second-order linear differential equation with constant coefficients. To summarize what we have accomplished: We successfully found the differential equation from its given solution by differentiating twice and eliminating the arbitrary constants. The process involved applying the product rule systematically and making strategic substitutions. Our final answer is d squared y dx squared minus 2 dy dx plus 2 y equals zero. This systematic approach can be applied to find differential equations from any given solution form.