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

Answers (1)

Answer      :  y=e^{-2 x}+c e^{-3 x}

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

Given         :  \frac{d y}{d x}+3 y=e^{-2 x}

Solution   : put in form \frac{d y}{d x}+P y=Q

\frac{d y}{d x}+3 y=e^{-2 x}

Step : 2

Find P and Q by comparing , P=3, Q=e^{-2 x}

Step : 3

Find integrating factor

\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\text { I.F }=e^{\int 3 d x} \\ &I . F=e^{3 x} \end{aligned}

Step : 4

Solution of equation

y \times I . F=\int Q \times I . F d x+c

Putting values

\begin{gathered} y \times e^{3 x}=\int e^{-2 x+3 x} d x+c \\ y e^{3 x}=\int e^{x} d x+c \\ y e^{3 x}=e^{x} d x+c \end{gathered}

divide by e^{3x}

y=e^{-2 x}+c e^{-3 x}

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