#### Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (iv) textbook solution.

Answer :  $\frac{1}{2}(\sin x-\cos x)+c e^{x}$

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

Given :  $\frac{d y}{d x}-y=\cos x$

Solution : differential equation is in form of

$\begin{gathered} \frac{d y}{d x}+P y=Q \\ P=-1 \quad, Q=\cos x \\ \text { I.F }=e^{\int P d x}=e^{\int-1 d x}=e^{-x} \end{gathered}$

Solution is

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

Integrate by $\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left\{f^{\prime}(x) \int g(x) d x\right\} d x$

$\begin{gathered} \text { Take } f(x)=\cos x \text { and } g(x)=e^{-x} \\ I=\cos x \int e^{-x} d x-\int\left\{(-\sin x) \int e^{-x} d x\right\} d x \\ I=-e^{-x} \cos x-\int-\sin x\left(-e^{-x}\right) d x \end{gathered}$

\begin{aligned} &I=-e^{-x} \cos x-\int e^{-x} \sin x d x \\ &I=-e^{-x} \cos x-\left\{\sin x \int e^{-x} d x-\int\left(\cos x \int e^{-x} d x\right) d x\right\} \end{aligned}

\begin{aligned} &I=-e^{-x} \cos x-\left(-e^{-x} \sin x\right)+\int-e^{-x} \cos x d x \\ &I=-e^{-x} \cos x+e^{-x} \sin x-\int e^{-x} \cos x d x \\ &I=e^{-x}(\sin x-\cos x)-I \\ &2 I=e^{-x}(\sin x-\cos x) \\ &I=\frac{e^{-x}}{2}(\sin x-\cos x) \end{aligned}

From (i)

\begin{aligned} &y e^{-x}=\int e^{-x} \cos x+c \\ &y e^{-x}=\frac{e^{-x}}{2}(\sin x-\cos x)+c \\ &y=\frac{1}{2}(\sin x-\cos x)+c e^{x} \end{aligned}

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