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Provide solution for RD Sharma class 12 chapter 21 Diffrential Equations excercise 21.4 question 8

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 y=e^{x}+e^{2 x} is the solution of given function


Differentiate the function


y=e^{x}+e^{2 x} is the given function                  


Differentiating with respect to  x

\Rightarrow \frac{d y}{d x}=e^{x}+e^{2 x} \cdots(i)

Differentiating with respect to  x

\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(e^{x}+2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+2\left(2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+4 e^{2 x} \cdots(i i) \end{aligned}

Put value (i) and (ii) in different equations

\begin{aligned} \frac{d^{2} y}{d x^{2}} &-3 \frac{d y}{d x}+2 y=0 \\ L H S &=\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y \\ &=\left(e^{x}+4 e^{2 x}\right)-3\left(e^{x}+2 e^{2 x}\right)+2\left(e^{x}+e^{2 x}\right) \\ &=e^{x}+4 e^{2 x}-3 e^{x}-6 e^{2 x}+2 e^{x}+2 e^{2 x} \\ &=0 \\ &=R H S \end{aligned}



y=e^{x}+e^{2 x} satisfies the diffrential equation

Now,when x=0

\begin{aligned} \mathrm{y} &=e^{0}+e^{2(0)} \\ &=1+e^{0} \\ &=1+1 \\ &=2 \end{aligned}

Now,when x=0

\begin{aligned} y^{\prime} &=e^{0}+2 e^{2(0)} \\ &=1+2(1) \\ y^{\prime} &=3 \end{aligned}

Thus, y(0)=2 \text { and } y^{\prime}(0)=3 satisfies the initial value problem

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