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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (i) textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (i) textbook solution.

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Answer : y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.

Give : \frac{d y}{d x}+3 y=e^{m x}, m is a given real number.

Hint: Use \int e^{x}dx

Explanation : \frac{d y}{d x}+3 y=e^{m x}

                     \frac{d y}{d x}+(3) y=e^{m x}

This is a first order linear differential equation of the form

                 =\frac{d y}{d x}+Py=Q

Here P=3 and Q=e^{mx}

The integrating factor If  of the differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 3 d x} \\ &=e^{3 \int d x} \\ &=e^{3 x} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int d c=x+C\right] \end{aligned}

Hence, the solution of differenttial equation is

\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x} e^{3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x+3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{x(m+3)} d x+C \end{aligned}

Case 1: m+3=0 \; or \; m=-3

When m+3=0, we have e^{x(m+3)}

              \begin{aligned} &=e^{0}=1 \\ &\Rightarrow y e^{3 x}=\int d x+C \\ &\Rightarrow y e^{3 x}=x+C \end{aligned}

              \begin{aligned} &\Rightarrow y e^{3 x} e^{-3 x}=(x+C) e^{-3 x} \\ &\Rightarrow y e^{3 x-3 x}=(x+C) e^{-3 x} \end{aligned}

              \Rightarrow y=(x+C) e^{-3 x} \quad\left[e^{3 x-3 x}=e^{0}=1\right]

Case 2: m+3 \neq 0 \text { or } m \neq-3

When m+3 \neq 0 we have

           \begin{aligned} &y e^{3 x}=\int e^{x(m+3)} d x+C \\ &\Rightarrow y e^{3 x}=\frac{e^{(m+3) x}}{m+3}+C \\ &\Rightarrow y e^{3 x} e^{-3 x}=\left(\frac{e^{(m+3) x}}{m+3}+C\right) e^{-3 x} \end{aligned}

           \begin{aligned} &\Rightarrow y e^{3 x} e^{-3 x}=\frac{\left(e^{m x} e^{3 x}\right) e^{-3 x}}{m+3}+C e^{-3 x} \\ &\Rightarrow y=\frac{e^{m x}}{m+3}+C e^{-3 x} \end{aligned}

Thus the solution of the given differential equation is

y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.

Posted by

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