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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 34 Maths Textbook Solution.

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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{d y}{d x}+P y=Q  where P,Q  are constants.

Give:  \begin{aligned} & \frac{d y}{d x}+2 y=x e^{4 x} \\ & \end{aligned}

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

\begin{aligned} &=\frac{d y}{d x}+P y=Q \\ &P=2, Q=x e^{4 x} \end{aligned}


If  of differential equation is

\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}

=e^{\int 2 d x} \\

=e^{2 \int d x} \\

=e^{2 x} \\

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

\begin{aligned} &\quad=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] \\ & \end{aligned}

=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 \\

=y e^{2 x}=e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C

\begin{aligned} &=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}

=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}

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