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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 15 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 15 textbook solution.

Answers (1)

Answer  :   \text { (b) } g(x)+\log (y-g(x)+1)=C

Hint: If the Differential equation is linear in y i.e. \frac{d y}{d x}+P(x) y=Q(x) \text { then } I F=e^{\int P(x) d x}

Given  :  \frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)

Explanation: The given differential equation is linear in y

\begin{aligned} &\frac{d y}{d x}+P(x) y=Q(x) \text { then IF }=e^{\int P(x) d x} \\ &I F=e^{\int g^{\prime}(x) d x} \\ &=e^{g(x)} \end{aligned}

General solution of the given differential equation,

\begin{aligned} &y e^{g(x)}=\int g(x) e^{g(x)} g^{\prime}(x) d x+C_{1} \\ &\text { Let } g(x)=t \Rightarrow g^{\prime}(x) d x=d t \end{aligned}

\begin{aligned} &y e^{g(x)}=\int t e^{t} d t+C_{1}\\ &y e^{g(x)}=t e^{t}-e^{t}+C_{1}\\ &y e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+C_{1}\\ &\text { Divide by } e^{g(x)} \end{aligned}

\begin{aligned} &y=g(x)-1+\frac{C_{1}}{e^{g(x)}} \\ &y=g(x)-1+C_{1} e^{-g(x)} \\ &y-g(x)+1=C_{1} e^{-g(x)} \end{aligned}

Taking logarithm on both sides

\begin{aligned} \log (y-g(x)+1) &=\log \left(C_{1} e^{-g(x)}\right) \\ &=\log C_{1}+\log e^{-g(x)} \\ &=\log C_{1}-g(x) \log e \\ &=\log C_{1}-g(x) \end{aligned}

\text { Hence, } g(x)+\log (y-g(x)+1)=C \quad \text { [ taking } \log C_{1}=C \text { ] }

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