Question

show that y=e^ax is the general solution of X dy/dx=y.log y​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\green{x\,\dfrac{dy}{dx}=y\cdot\,log(y)}}$

$\sf{\implies\,\dfrac{dy}{y\cdot\,log(y)}=\dfrac{dx}{x}}$

$\sf{\implies\,\displaystyle\int\dfrac{dy}{y\cdot\,log(y)}=\int\dfrac{dx}{x}}$

$\rm{\purple{Put\,\,\,\,\,\,log(y)=v}}$

$\rm{\implies\purple{\dfrac{dy}{y}=dv}}$

So,

$\sf{\implies\,\displaystyle\int\dfrac{dv}{v}=\int\dfrac{dx}{x}}$

$\sf{\implies\,ln|v|=ln|x|+c}$

Let c = ln(a),

$\sf{\implies\,ln|v|=ln|x|+ln|a|}$

$\sf{\implies\,ln|v|=ln|ax|}$

$\sf{\implies\,v=ax}$

$\sf{\implies\,log(y)=ax}$

$\sf{\implies\,y=e^{ax}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = {e}^{ax}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}{e}^{ax}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{ax}\dfrac{d}{dx}(ax)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =ay\dfrac{d}{dx}(x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =ay \times 1$

$\rm :\longmapsto\:\dfrac{dy}{dx} =ay$

Now, Consider,

$\rm :\longmapsto\:x\dfrac{dy}{dx}$

$\rm \:  =  \: xay$

can be rewritten as

$\rm \:  =  \: y(ax)$

which can be rewritten as

$\rm \:  =  \: y log{e}^{ax}$

$\: \: \: \bf \: [ \because \: log {e}^{x} = x \: ]$

$\rm \:  =  \: ylogy$

Hence,

$\rm \implies\:\boxed{ \tt{ \: x \: \dfrac{dy}{dx} \: = \: y \: logy \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$