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#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 53 maths textbook solution.

Answer : $\frac{1}{41}(4 \sin 4 x+5 \cos 4 x)+c e^{-5 x}$

Hint        : You must know the rules of solving differential equation and integration.

Given      : $\frac{dy}{dx}+5y=\cos 4x$

Solution  : $\frac{dy}{dx}+5y=\cos 4x$

Compare with,

$\frac{dy}{dx}+Py=Q$

When $P=5, Q= \cos 4x$

Therefore,

\begin{aligned} I . F &=e^{\int P d x} \\ &=e^{\int 5 d x} \\ &=e^{5 x} \end{aligned}

The solution is ,

\begin{aligned} y \times I . F &=\int(I . F \times Q) d x+c \\ y e^{5 x} &=\int e^{5 x} \times \cos 4 x d x+c \\ y e^{x} &=I+c \end{aligned}         ....(i)

Where,

\begin{aligned} I &=\int e^{5 x} \cos 4 x d x \\ I &=e^{5 x} \int \cos 4 x d x-\int\left[\frac{d e^{5 x}}{d x} \int \cos 4 x d x\right] d x \\ I &=e^{5 x} \frac{\sin 4 x}{4}-\frac{5}{4} \int e^{5 x} \sin 4 x d x \end{aligned}

\begin{aligned} &I=\frac{e^{5 x} \sin 4 x}{4}-\frac{5}{4}\left[e^{5 x} \int \sin 4 x d x-\int\left\{\frac{d e^{5 x}}{d x} \int \sin 4 x d x\right\} d x\right] \\ &I=\frac{e^{5 x} \sin 4 x}{4}-\frac{5}{4}\left[\frac{-e^{5 x} \cos 4 x}{4}+\frac{5}{4} \int e^{5 x} \cos 4 x d x\right] \\ &I=\frac{e^{5 x} \sin 4 x}{4}+\frac{5}{16} e^{5 x} \cos 4 x-\frac{25}{16} \int e^{5 x} \cos 4 x d x \end{aligned}

$\begin{gathered} I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x)-\frac{25}{16} I \\ \frac{25}{16} I+I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x) \\ \frac{41}{16} I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x) \\ I=\frac{e^{5 x}}{41}(4 \sin 4 x+5 \cos 4 x) \end{gathered}$

Therefore, required solution is

\begin{aligned} y e^{5 x} &=\frac{e^{5 x}}{41}(4 \sin 4 x+5 \cos 4 x)+c \\ y &=\frac{1}{41}(4 \sin 4 x+5 \cos 4 x)+c e^{-5 x} \end{aligned}