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Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 21 sub question (iii)

Answers (1)

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Answer:

y=\frac{a}{x+a}  is the function which satisfies the differential equation

Hint:

Just differentiate the function and substitute in given equation

Given:

y=\frac{a}{x+a} 

Solution:

Differentiating on both sides with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{(x+a)(0)-a(1)}{(x+a)^{2}} \\\\ &\frac{d y}{d x}=\frac{-a}{(x+a)^{2}} \end{aligned}                                .............(i)

Substitute equation (i) in differential equation

\begin{aligned} &x \frac{d y}{d x}+y=y^{2} \\\\ &L H S=x \frac{d y}{d x}+y \end{aligned}

            \begin{aligned} &=x\left(\frac{-a}{(x+a)^{2}}\right)+y \\\\ &=\frac{-x a}{(x+a)^{2}}+\frac{a}{x+a} \\\\ &=\frac{-x a+a(x+a)}{(x+a)^{2}} \end{aligned}

            \begin{aligned} &=\frac{-x a+x a+a^{2}}{(x+a)^{2}} \\\\ &=\frac{a^{2}}{(x+a)^{2}} \\\\ &=\left(\frac{a}{(x+a)}\right)^{2} \end{aligned}

            =y^{2}                                                ..................(ii)

LHS = RHS

Thus, y=\frac{a}{x+a}  is the function which satisfies the differential equation.

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