Question

Solution of the differential equation \(\Big[\frac{e^{-2\sqrt{x}}}{\sqrt{x}} – \frac{y}{\sqrt{x}}\Big]  \frac{dy}{dx} = 1, \,x≠0\) is _______

(a) \(ye^{2\sqrt{x}}=2\sqrt{x}+c\)

(b) \(ye^{-2\sqrt{x}}=2x^{-\frac{3}{2}}+c\)

(c) \(ye^{-2\sqrt{x}}=3\sqrt{x}+c\)

(d) \(ye^{2\sqrt{x}}=3x^{\frac{3}{2}}+c\)

The question was posed to me in my homework.

My question is based upon First Order Linear Differential Equations topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right option is (a) \(ye^{2\sqrt{x}}=2\sqrt{x}+c\)

For explanation I would say: \( \frac{1}{\sqrt{x}} e^{-2\sqrt{x}} – \frac{1}{\sqrt{x}} y = \frac{dy}{dx} \rightarrow \frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{1}{\sqrt{x}} e^{-2\sqrt{x}}\) is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x only.

given DE is linear DE in y here \(P=\frac{1}{\sqrt{x}}, Q=\frac{1}{\sqrt{x}} e^{-2\sqrt{x}}\)

\(e^{\int P \,dx} = e^{∫ \frac{1}{\sqrt{x}} \,dx} = e^{2\sqrt{x}},\)

Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)

\(ye^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} e^{-2\sqrt{x}} e^{2\sqrt{x}} \,dx + c = 2\sqrt{x} + c\) is the solution.