# 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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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.