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

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

Answers (1)

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}

Posted by

infoexpert23

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