question_answer

Solve the differential equation

\[y{{e}^{y}}dx=({{y}^{3}}+2x{{e}^{y}})\,dy,\] y(0) = 1

OR

Find the differential equation of the family of curves \[y={{e}^{2x}}(acos2x+b\sin 2x),\] where a and b are arbitrary constants.

Answer:

We have, \[y{{e}^{y}}dx=({{y}^{3}}+2x{{e}^{y}})\,dy\]

\[\Rightarrow \]   \[\frac{dx}{dy}=\frac{{{y}^{3}}+2x{{e}^{y}}}{y{{e}^{y}}}\]

\[\Rightarrow \]   \[\frac{dx}{dy}={{y}^{2}}{{e}^{-y}}+\frac{2x}{y}\]

\[\Rightarrow \]   \[\frac{dx}{dy}-\frac{2}{y}x={{y}^{2}}{{e}^{-y}}\]

which is a linear differential equation.

\[\therefore \]      \[IF={{e}^{\int{-2/y\,\,dy}}}={{e}^{-2\,\,\log \,\,y}}\]

\[={{e}^{\log \,\,{{y}^{-2}}}}={{y}^{-2}}\]

\[\therefore \] Solution is given by

\[x\cdot {{y}^{-\,2}}=\int{{{y}^{2}}{{e}^{-y}}\cdot {{y}^{-2}}dy+C}\]

\[\Rightarrow \]   \[x{{y}^{-\,2}}=\int{{{e}^{-y}}dy+C}\]

\[\Rightarrow \]   \[\frac{x}{{{y}^{2}}}=-{{e}^{-y}}+C\]                                 ?(i)

It is given that y(0) = 1, i.e. y = 1 when x = 0.

Putting x = 0 and y = 1 in Eq. (i), we get

\[0=-\,{{e}^{-1}}+C\] \[\Rightarrow \] \[C=\frac{1}{e}\]

Putting \[C=\frac{1}{e}\] in Eq. (i), we get

\[\frac{x}{{{y}^{2}}}=-\,{{e}^{-y}}+\frac{1}{e}\]

\[\Rightarrow \]   \[x={{y}^{2}}({{e}^{-1}}-{{e}^{-y}})\]

Hence, \[x={{y}^{2}}({{e}^{-1}}-{{e}^{-y}})\] is the required solution.

OR

Given equation of family of curves is

\[y={{e}^{2x}}(a\cos 2x+b\sin 2x)\]        ?(i)

Since, there are two arbitrary constants, therefore we will differentiate it two times w.r.t. x.

On differentiating Eq. (i), w.r.t. x, we get

\[\frac{dy}{dx}={{e}^{2x}}(-\,a\sin 2x\cdot 2+b\cos 2x\cdot 2)\]

\[+\,\,(a\cos 2x+b\sin 2x)\cdot {{e}^{2x}}\cdot 2\]

\[\Rightarrow \]   \[\frac{dy}{dx}=2{{e}^{2x}}(-a\sin 2x+b\cos 2x)+2y\]

[using Eq.. (i)]

\[\Rightarrow \]   \[\frac{dy}{dx}-2y=2{{e}^{2x}}(-a\sin 2x+b\cos 2x)\]                                                      ?(ii)

Again, differentiating Eq. (ii). w.r.t. x, we get

\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}=2\] \[\left[ \begin{align}   & {{e}^{2x}}(-a\cos 2x\cdot 2-b\sin 2x\cdot 2) \\  & +(-a\sin 2x+b\cos 2x)\cdot {{e}^{2x}}\cdot 2 \\ \end{align} \right]\]

\[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}=2\] \[\left[ \begin{align}   & -2{{e}^{2x}}(a\cos 2x+b\sin 2x) \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( \frac{dy}{dx}-2y \right) \\ \end{align} \right]\]

[using Eq. (ii)]

\[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}=2\left[ -\,2y+\frac{dy}{dx}-2y \right]\]

[using Eq. (i)]

\[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}=-\,8y+2\frac{dy}{dx}\]

\[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-4\frac{dy}{dx}+8y=0\]

which is the required differential equation.