Answer to Question #139609 in Differential Equations for LAVANYA

Question #139609
p^2(x^2-a^2)-2xyp+y^2+a^4=0 find G.S and S.S
1
Expert's answer
2020-10-28T17:41:00-0400

Given differential equation is, "p^2(x^2-a^2)-2xyp+y^2+a^4=0"

I can write it as,

"p^2x^2-2xyp+y^2=a^2p^2 - a^4"


"\\implies (y-xp)^2 = a^2(p^2-a^2)"


Then, "(y-xp) = \\sqrt{a^2(p^2-a^2)}"

"y=xp \\pm \\sqrt{a^2(p^2-a^2)}"



This equation is same as charpit's equation, "y = xf(p)+\\phi(p)"

So general solution of the equation will be,

"y=cx \\pm \\sqrt{a^2(c^2-a^2)}"

where c is any constant.


For particular solution, we need some boundary conditions.

Let x = 0 at y=0

then,

"0=0c \\pm \\sqrt{a^2(c^2-a^2)} \\implies c = \\pm a"

so, particular solution will be,

"y=\\pm ax"




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