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Question
Solve the following differential equation:
y2 dx + (xy + x2)dy = 0
Solution
y2 dx + (xy + x2)dy = 0
∴ (xy + x2)dy = - y2 dx
∴ `"dy"/"dx" = (- "y"^2)/("xy + x"^2)`
Put y = vx
∴ `"dy"/"dx" = "v + x""dv"/"dx"`
Substituting these values in (1), we get
`"v + x""dv"/"dx" = (- "v"^2"x"^2)/("x" * "vx + x"^2) = (- "v"^2)/("v + 1")`
∴ `"x" "dv"/"dx" = (- "v"^2)/("v + 1") - "v" = (- "v"^2 - "v"^2 - "v")/("v + 1")`
∴ `"x" "dv"/"dx" = (- 2"v"^2 - "v")/("v + 1") = -(("2v"^2 + "v")/("v + 1"))`
∴ `("v + 1")/("2v"^2 + "v")"dv" = - 1/"x" "dx"`
Integrating both sides, we get
`int ("v + 1")/("2v"^2 + "v")"dv" = - int 1/"x" "dx"`
∴`int ("v + 1")/("v"("2v" + 1))"dv" = - int 1/"x" "dx"`
∴ `int(("2v" + 1) - "v")/("v"("2v" + 1))"dv" = - int 1/"x" "dx"`
∴ `int(1/"v" - 1/("2v + 1"))"dv" = - int 1/"x" "dx"`
∴ `int 1/"v" "dv" - int1/("2v + 1")"dv" = - int 1/"x" "dx"`
∴ `log |"v"| - 1/2 log |2"v" + 1| = - log |"x"| + log "c"`
∴ `2 log |"v"| - log |2"v" + 1| = - 2 log |"x"| + 2 log "c"`
∴ `log |"v"^2| - log |"2v" + 1| = - log |"x"^2| + log "c"^2`
∴ `log |"v"^2/("2v" + 1)| = log |"c"^2/"x"^2|`
∴ `"v"^2/("2v" + 1) = "c"^2/"x"^2`
∴ `("y"^2/"x"^2)/(2 ("y"/"x") + 1) = "c"^2/"x"^2`
∴ `"y"^2/("x"("2y + x")) = "c"^2/"x"^2`
∴ xy2 = c2(x + 2y)
This is the general solution.
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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)
