The general solution of cos ^2 x frac{d y}{d | vedclass

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Question

(A) Given the differential equation: $\cos ^2 x \frac{d y}{d x}+y=	an x$. Divide by $\cos ^2 x$: $\frac{d y}{d x} + y \sec ^2 x = 	an x \sec ^2 x$.

Vedclass · Accepted Answer

$y e^{	an x}=(	an x-1) e^{	an x}+c$

Vedclass · Answer

(A) Given the differential equation: $\cos ^2 x \frac{d y}{d x}+y=	an x$. Divide by $\cos ^2 x$: $\frac{d y}{d x} + y \sec ^2 x = 	an x \sec ^2 x$. This is a linear differential equation of the form $\frac{d y}{d x} + P(x)y = Q(x)$,where $P(x) = \sec ^2 x$ and $Q(x) = 	an x \sec ^2 x$. The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int \sec ^2 x dx} = e^{	an x}$. The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$. $y e^{	an x} = \int 	an x \sec ^2 x e^{	an x} dx + C$. Let $u = 	an x$,then $du = \sec ^2 x dx$. The integral becomes $\int u e^u du$. Using integration by parts: $\int u e^u du = u e^u - \int e^u du = u e^u - e^u = e^u(u-1)$. Substituting back: $y e^{	an x} = e^{	an x}(	an x - 1) + C$.

The general solution of $\cos ^2 x \frac{d y}{d x}+y=\tan x$ is

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