Question

If y = xe^xy then prove that dy/dX = y(1 + xy)/ x(1 - xy)​

Answer 1

SOLUTION

GIVEN

$\sf{y = x \: {e}^{xy} \: }$

TO PROVE

$\displaystyle \sf{ \frac{dy}{dx} = \frac{y(1 + xy)}{x(1 - xy)} \: }$

PROOF

Here it is given that

$\sf{y = x \: {e}^{xy} \: } \: \: ...(1)$

$\displaystyle \sf{ {e}^{xy} = \frac{y}{x} } \: \: \: ....(2)$

Differentiating both sides of Equation (1) with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \: {e}^{xy} + x \: {e}^{xy} \bigg( y + x \frac{dy}{dx} \bigg) \: }$

$\implies \displaystyle \sf{ \frac{dy}{dx} = \frac{y}{x} + x \times \frac{y}{x} \times \bigg( y + x \frac{dy}{dx} \bigg) \: }$ ( by Eqn. 2)

$\implies \displaystyle \sf{ x\frac{dy}{dx} = y + x y \bigg( y + x \frac{dy}{dx} \bigg) \: }$

$\implies \displaystyle \sf{ x\frac{dy}{dx} = y + x {y}^{2} + {x}^{2}y \frac{dy}{dx} \: }$

$\implies \displaystyle \sf{ x\frac{dy}{dx} - {x}^{2}y \frac{dy}{dx} = y + x {y}^{2} \: }$

$\implies \displaystyle \sf{ \big(x - {x}^{2} y \big)\frac{dy}{dx}= y + x {y}^{2} \: }$

$\implies \displaystyle \sf{ \frac{dy}{dx}= \frac{y + x {y}^{2} }{x - {x}^{2} y} \: }$

$\implies \displaystyle \sf{ \frac{dy}{dx}= \frac{y(1 + x {y}^{} )}{x(1 - {x}^{} y)} \: }$

Hence proved

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Answer 2

Step-by-step explanation: