Find the particular solution of the differential equation \frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}.

Category: QuestionsFind the particular solution of the differential equation \frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}.
Editor">Editor Staff asked 11 months ago

Find the particular solution of the differential equation \frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}.
 
(a) (log⁡y)^2+(log⁡x)^2=0
 
(b) (log⁡y)^2-(log⁡x)^2=0
 
(c) log⁡y-log⁡x=0
 
(d) 2 log⁡x+log⁡y=0
 
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Enquiry is from Methods of Solving First Order & First Degree Differential Equations in division Differential Equations of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Right choice is (b) (log⁡y)^2-(log⁡x)^2=0
 
Easy explanation: \frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}
 
Separating the variables, we get
 
\frac{5 \,log⁡y}{y} dy=\frac{9 \,log⁡x}{x} dx
 
Integrating both sides, we get
 
5\int \frac{log⁡y}{y} \,dy=9\int \frac{log⁡x}{x} \,dx –(1)
 
First, for integrating \frac{log⁡y}{y}
 
Let log⁡y=t
 
Differentiating w.r.t y, we get
 
\frac{1}{y} dy=dt
 
∴\int \frac{log⁡y}{y} \,dy=\int t \,dt
 
=\frac{t^2}{2}=\frac{log⁡y^2}{2}
 
Similarly integrating \frac{log⁡x}{x}
 
Let log⁡x=t
 
Differentiating w.r.t x, we get
 
\frac{1}{x} dx=dt
 
∴\int \frac{log⁡x}{x} dy=\int t \,dt
 
=\frac{t^2}{2}=\frac{(log⁡x)^2}{2}
 
Hence, equation (1), becomes
 
\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C –(2)
 
Given y=2, we get x=2
 
Substituting the values in equation (2), we get
 
\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C
 
C=0
 
Therefore, the equation becomes (log⁡y)^2=(log⁡x^2)
 
∴(log⁡y)^2-(log⁡x)^2=0.