The general solution of the differential equation y(x^{2}y+e^{x})dx-e^{x}dy=0 is

  • Option 1)

    x^{3}y-3e^{x}= cy

  • Option 2)

    x^{3}y+3e^{x}= 3cy

  • Option 3)

    y^{3}x-3e^{y}= cx

  • Option 4)

    y^{3}x+3e^{y}= cx

 

Answers (1)

 

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

 y(x^{2}y+e^{x})dx=e^{x}dy

\frac{1}{y^{2}}\frac{dy}{dx}-\frac{1}{y}=\frac{x^{2}}{e^{x}}

Let\:\:\frac{1}{y}=t

\therefore \frac{-1}{y^{2}}\frac{dy}{dx}=\frac{dt}{dx}

\therefore \frac{dt}{dx}+t=\frac{-x^{2}}{e^{x}}

\int 1.dx=x

if\:\because e^{x}

Solution is

t.e^{x}=\int \frac{-x^{2}}{e^{x}}\times e^{x}dx

=\frac{-x^{3}}{3}+C

\frac{1}{y}e^{x}=\frac{-x^{3}}{3}+C

\therefore 3e^{x}+x^{3}y=3cy


Option 1)

x^{3}y-3e^{x}= cy

Option is incorrect

Option 2)

x^{3}y+3e^{x}= 3cy

Option is correct

Option 3)

y^{3}x-3e^{y}= cx

Option is incorrect

Option 4)

y^{3}x+3e^{y}= cx

Option is incorrect

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