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.
You need to login to perform this action.
You will be redirected in
3 sec