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  • Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 9 maths textbook solution

Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 9 maths textbook solution

Answers (1)

Answer:

 \frac{1}{\sqrt{x}}

Hint:

 Use the concept of linear Differential equation is of the form

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

Given:

 \left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right )\frac{\mathrm{d} x}{\mathrm{d} y}=1,\quad x\neq 0

Solution:

\begin{aligned} &\Rightarrow \left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right )\frac{\mathrm{d} x}{\mathrm{d} y}=1 \\ &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right ) \\ &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}} \end{aligned}

The equation in the form

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

Where

\begin{aligned} &P=\frac{1}{\sqrt{x}} \end{aligned}

So, the answer is

\begin{aligned} &P=\frac{1}{\sqrt{x}} \end{aligned}

\text { Note: In question there will be } \frac{\mathrm{d} x}{\mathrm{d} y} \text { in place of } \frac{\mathrm{d} y}{\mathrm{d} x}.

Posted by

Gurleen Kaur

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