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  • Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.3 question 12 maths

Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.3 question 12 maths

Answers (1)

Answer:

y=e^{x}(A \cos x+B \sin x)  is a solution of given equation

Hint:

Differentiate the given solution and substitute in the differential equation.

Given:

y=e^{x}(A \cos x+B \sin x)  is a solution of the equation


Solution:

Differentiating on both sides with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[e^{x}(A \cos x+B \sin x)\right] \\\\ &\frac{d y}{d x}=\frac{d}{d x}\left[e^{x} A \cos x\right]+\frac{d}{d x}\left[e^{x} B \sin x\right] \end{aligned}

\frac{d y}{d x}=A\left[e^{x} \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}\left(e^{x}\right)\right]+B\left[e^{x} \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}\left(e^{x}\right)\right]

\begin{aligned} &\frac{d y}{d x}=A\left[e^{x}(-\sin x)+\cos x\left(e^{x}\right)\right]+B\left[e^{x}(\cos x)+\sin x\left(e^{x}\right)\right] \\\\ &\frac{d y}{d x}=e^{x}[-A \sin x+A \cos x]+e^{x}[B \cos x+B \sin x] \end{aligned}            ............(i)

Differentiate equation (i) with respect to x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[e^{x}[-A \sin x+A \cos x]+e^{x}[B \cos x+B \sin x]\right] \\\\ &\frac{d^{2} y}{d x^{2}}=A \frac{d}{d x}\left[-e^{x} \sin x+e^{x} \cos x\right]+B \frac{d}{d x}\left[e^{x} \cos x+e^{x} \sin x\right] \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=A\left[-\left(e^{x} \cos x+e^{x} \sin x\right)+e^{x}(-\sin x)+e^{x} \cos x\right]+B\left[\left(e^{x}(-\sin x)+e^{x} \cos x\right)+e^{x}(\cos x)+e^{x} \sin x\right] \\\\ &\frac{d^{2} y}{d x^{2}}=A\left[e^{x}(-\cos x-\sin x)+e^{x}(\cos x-\sin x)\right]+B\left[e^{x}(\cos x-\sin x)+e^{x}(\cos x+\sin x)\right] \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=e^{x}[-A \sin x-A \cos x+A \cos x-A \sin x]+e^{x}[B \cos x-B \sin x+B \cos x+B \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=e^{x}[-2 A \sin x]+e^{x}[2 B \cos x] \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=e^{x}[2 B \cos x-2 A \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=2 e^{x}[-A \sin x+B \cos x]+2 e^{x}[A \cos x+B \sin x]-2 e^{x}[A \cos x+B \sin x] \end{aligned}

                                  [on Adding and subtracting the value of y in the above step ]

\frac{d^{2} y}{d x^{2}}=2\left[e^{x}(-A \sin x+B \cos x)+e^{x}(A \cos x+B \sin x)\right]-2 e^{x}\left ( A\cos x+B\sin x \right )

  Put value of y as per question and value of equation (i) in above equation                                                                                            

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=2 \frac{d y}{d x}-2 y \\\\ &\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0 \end{aligned}

Hence it is proved that y=e^{x}(A \cos x+B \sin x) is solution of given equation.

 

Posted by

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