Gmath

2yxe^{x/y} dx + (y-2xe^{x/y} dx = 0

Question 2:

Show that differential equation $\displaystyle 2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0$ is homogeneous and find its particular solution, given that x = 0 when y = 1. [CBSE(North) 2016; (south) 2016, 2017]

Solution:

Given: $\displaystyle 2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0$

$\displaystyle \Rightarrow \dfrac{dx}{dy} = -\dfrac{y - 2x e^{x/y}}{2y e^{x/y}}$

$\displaystyle \Rightarrow \dfrac{dx}{dy} = \dfrac{2x e^{x/y}-y}{2y e^{x/y}}$

Let, $\displaystyle F(x , y) = \dfrac{2x e^{x/y}-y}{2y e^{x/y}}$

∴ $\displaystyle F(\lambda x , \lambda y)= \dfrac{2\lambda x e^{\lambda x/\lambda y}-\lambda y}{2\lambda y e^{\lambda x/\lambda y}}$

$\displaystyle \Rightarrow F(\lambda x , \lambda y) = \lambda^0\dfrac{2x e^{x/y}-y}{2y e^{x/y}}$

$\displaystyle \Rightarrow F(\lambda x , \lambda y) = \lambda^0 F(x , y)$

Hence, given differential equation is homogeneous.

Now,

$\displaystyle \dfrac{dx}{dy} = \dfrac{2x e^{x/y}-y}{2y e^{x/y}}$ . . . . . . . . (i)

Let, z = vy

$\Rightarrow \dfrac{dx}{dy} = v + \dfrac{dv}{dy}$

From (i)

$\displaystyle v + \dfrac{dv}{dy} = \dfrac{2v y e^{v y/y}-y}{2y e^{v y/y}}$

$\displaystyle \Rightarrow \dfrac{dv}{dy} = \dfrac{2v e^{v}-1}{2 e^{v}} - v$

$\displaystyle \Rightarrow \dfrac{d v}{dy} = \dfrac{2v e^{v}-1 - 2v e^v}{2 e^{v}}$ `

$\displaystyle \Rightarrow \dfrac{d v}{dy} = \dfrac{-1 }{2 e^{v}}$

$\displaystyle \Rightarrow 2 e^{v} dv = - dy$

$\displaystyle \Rightarrow \int 2 e^{v} dv = - \int dy$

$\displaystyle \Rightarrow 2 e^v = - \log y + C$

$\displaystyle \Rightarrow 2 e^{x/y} + \log y = C$

When x = 0, y = 1

∴ $2e^0 + \log 1 = C$

$\Rightarrow C = 2$

Hence the required solution is

$\displaystyle 2 e^{x/y} + \log y = 2$

Some other question

Question 1: Solve the following differential equation: $(1 + e^{y/x})dy + e^{y/x}(1 - \frac{y}{x}) = 0,$ x ≠ 0

Solution : For solution click here

Question 3: show that the given differential equation is homogeneous and solve it.

$\displaystyle\{x\cos (\frac{y}{x})+y\sin (\frac{y}{x})\}y dx = \{y\sin (\frac{y}{x}) - x\cos (\frac{y}{x})\}x dy$ .

Solution: For solution click here

Team Gmath

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