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Question
Find the particular solution of the differential equation:
2y ex/y dx + (y - 2x ex/y) dy = 0 given that x = 0 when y = 1.
Solution
`2ye^(x/y)dx+(y-2xe^(x/y))dy=0`
`=>dx/dy=(2xe^(x/y-y))/(2ye^(x/y))`
Given differential equation is a homogeneous differential equation.
∴ Put x = vy
`dx/dy=v+y (dv)/dy`
`v+y(dv)/dy=(2ve^v-1)/(2e^v)`
`=>y(dv)/dy=(2ve^v-1)/(2e^v)-v`
`=>y(dv)/dy=-1/(2e^v)`
`=>2e^vdv=-1/ydy`
Integrating on both the sides
`=>2inte^vdv=-int1/ydy`
`=>2e^v=-log|y|+logC`
`=>2e^v=log|c/y|`
`=>2e^(x/y)=log|c/y|`
Given that at x = 0, y = 1
`2e^0= log|c/1|`
⇒ C = e2
`:.2e^(x/y)=log""e^2/y`
`=>logy=-2e^(x/y)+2`
`=>y=e^2-2e^(x/y)`
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|
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
Based on the above, answer the following questions:
- Show that (x2 – y2) dx + 2xy dy = 0 is a differential equation of the type `dy/dx = g(y/x)`. (2)
- Solve the above equation to find its general solution. (2)
