Question:

Solve the differential equation \([\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}]\frac{dx}{dy}=1(x≠0)\)

Updated On: Sep 21, 2023
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Solution and Explanation

The correct answer is: \(ye^{2\sqrt{x}}=2\sqrt{x}+C\)
\([\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}]\frac{dx}{dy}=1\)
\(\implies \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\)
\(⇒\frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)
This equation is a linear differential equation of the form
\(\frac{dy}{dx}+py=Q\),where \(p=\frac{1}{\sqrt{x}}\) and \(Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.\)
Now,\(I.F.=e^{∫pdx}=e^{∫\frac{1}{\sqrt{x}}dx}=e^{2\sqrt{x}}\)
The general solution of the given differential equation is given by,
\(y(I.F.)=∫(Q×I.F.)dx+C\)
\(⇒ye^{2\sqrt{x}}=∫(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}×e^{2\sqrt{x}})dx+C\)
\(⇒ye^{2\sqrt{x}}=∫\frac{1}{\sqrt{x}}dx+C\)
\(⇒ye^{2\sqrt{x}}=2\sqrt{x}+C\)
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