The solution of the equation frac{dy}{dx} + y | vedclass

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(A) This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = x^m \cos x$. First,we calculate the integr

Vedclass · Accepted Answer

$(m + 1)y = x^{m + 1} \cos x + c(m + 1) \cos x$

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(A) This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = x^m \cos x$. First,we calculate the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$. The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \cdot \sec x = \int (x^m \cos x) \cdot \sec x dx + c$. Since $\cos x \cdot \sec x = 1$,the equation simplifies to: $y \sec x = \int x^m dx + c$. Integrating $x^m$ with respect to $x$: $y \sec x = \frac{x^{m + 1}}{m + 1} + c$. Multiplying both sides by $(m + 1) \cos x$: $(m + 1) y = x^{m + 1} \cos x + c(m + 1) \cos x$.

The solution of the equation $\frac{dy}{dx} + y \tan x = x^m \cos x$ is

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