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Consider the differential equation, xdydx=y(log⁡ey−log⁡ex+1)x\frac{dy}{dx} = y(\log_e y - \log_e x + 1), then which of the following are true?

(A) It is a linear differential equation

(B) It is a homogenous differential equation

(C) Its general solution is log⁡e(yx)=Cx\log_e\left(\frac{y}{x}\right) = Cx, where C is constant of integration

(D) Its general solution is log⁡e(xy)=Cy\log_e\left(\frac{x}{y}\right) = Cy, where C is constant of integration

(E) If y(1)=1y(1) = 1, then its particular solution is y=xy = x

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

The differential equation is xdydx=y(log⁡ey−log⁡ex+1)x\frac{dy}{dx} = y(\log_e y - \log_e x + 1).

Rewrite using log⁡ey−log⁡ex=log⁡e(yx)\log_e y - \log_e x = \log_e\left(\frac{y}{x}\right):

xdydx=y(log⁡e(yx)+1)x\frac{dy}{dx} = y\left(\log_e\left(\frac{y}{x}\right) + 1\right)

dydx=yx(log⁡e(yx)+1)\frac{dy}{dx} = \frac{y}{x}\left(\log_e\left(\frac{y}{x}\right) + 1\right)


A linear differential equation has the form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x) where yy should not appear inside any function.

The equation contains log⁡ey\log_e y, so yy appears inside a logarithm.

Statement (A) is FALSE.


A homogeneous differential equation can be written as dydx=f(yx)\frac{dy}{dx} = f\left(\frac{y}{x}\right).

The equation dydx=yx(log⁡e(yx)+1)\frac{dy}{dx} = \frac{y}{x}\left(\log_e\left(\frac{y}{x}\right) + 1\right) is a function of yx\frac{y}{x} only.

Statement (B) is TRUE.


Let v=yxv = \frac{y}{x}, so y=vxy = vx and dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}.

v+xdvdx=v(log⁡ev+1)v + x\frac{dv}{dx} = v(\log_e v + 1)

xdvdx=vlog⁡ev+v−vx\frac{dv}{dx} = v\log_e v + v - v

xdvdx=vlog⁡evx\frac{dv}{dx} = v\log_e v

dvvlog⁡ev=dxx\frac{dv}{v\log_e v} = \frac{dx}{x}


Let u=log⁡evu = \log_e v, then du=dvvdu = \frac{dv}{v}.

duu=dxx\frac{du}{u} = \frac{dx}{x}

Integrating both sides:

log⁡e∣u∣=log⁡e∣x∣+log⁡e∣C∣\log_e|u| = \log_e|x| + \log_e|C|

u=Cxu = Cx


Since u=log⁡evu = \log_e v:

log⁡ev=Cx\log_e v = Cx

Since v=yxv = \frac{y}{x}:

log⁡e(yx)=Cx\log_e\left(\frac{y}{x}\right) = Cx

Statement (C) is TRUE.

Statement (D) is FALSE.


For the initial condition y(1)=1y(1) = 1:

log⁡e(11)=C(1)\log_e\left(\frac{1}{1}\right) = C(1)

log⁡e(1)=C\log_e(1) = C

0=C0 = C

So C=0C = 0:

log⁡e(yx)=0\log_e\left(\frac{y}{x}\right) = 0

yx=e0\frac{y}{x} = e^0

yx=1\frac{y}{x} = 1

y=xy = x

Statement (E) is TRUE.


True statements: (B), (C), and (E)

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