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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 34 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 34 textbook solution.

Answers (1)

Answer : y=\frac{x e^{4 x}}{6}-\frac{1}{36} e^{4 x}+C e^{-2 x}

Hint :To solve this equation we use \frac{dy}{dx}+Py=Q where P,Q are constants.

Give : \frac{d y}{d x}+2 y=x e^{4 x}

Solution : \frac{d y}{d x}+(2) y=x e^{4 x}

              =\frac{dy}{dx}+Py=Q

              P=2, Q=x e^{4 x}

If  of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 2 d x} \\ &=e^{2 \int d x} \\ &=e^{2 x} \end{aligned}

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(e^{2 x}\right)=\int x e^{4 x} e^{2 x} d x+C \\ &=y e^{2 x}=\int x e^{6 x} d x+C \end{aligned}

\begin{aligned} &=y e^{2 x}=\int x e^{6 x} d x+C\left[\int f(x) g(x) d c=f(x) \int[g(x) d x] \int f(x)\left[\int g(x) d x\right] d x\right] \\ &=y e^{2 x}=x\left(\int e^{6 x} d x\right)-\int \frac{d}{d x}(x)\left(\int e^{6 x} d x\right) d x+C \\ &=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\int 1\left(\frac{e^{6 x}}{6}\right) d x+C \\ &=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\frac{1}{6}\left(\frac{e^{6 x}}{6}\right)+C \end{aligned}

\begin{aligned} &=y e^{2 x}=e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C \\ &=y e^{2 x} e^{-2 x}=e^{-2 x}\left[e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C\right] \end{aligned}

\begin{aligned} &=y e^{2 x-2 x}=e^{-2 x} e^{6 x}\left(\frac{x}{6}\right)-e^{-2 x} \frac{1}{36}\left(e^{6 x}\right)+C e^{-2 x} \\ &=y e^{0}=e^{6 x-2 x}\left(\frac{x}{6}\right)-e^{6 x-2 x} \frac{1}{36}+C e^{-2 x} \\ &=y=e^{4 x}\left(\frac{x}{6}\right)-e^{4 x} \frac{1}{36}+C e^{-2 x} \end{aligned}

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