solve the differential equation.
dy/dx = xy / (√ x^2 -4 ).



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2 thoughts on “solve the differential equation.<br />dy/dx = xy / (√ x^2 -4 ).<br /><br /><br /><br />​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given Differential equation is,

   [tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{xy}{ \sqrt{ {x}^{2} – 4 }} [/tex]

   To solve this Differential equation, we use method of separated the variable.

   The given Differential equation can be rewritten as

   [tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x}{ \sqrt{ {x}^{2} – 4 }} \times y[/tex]

   [tex]\rm :\longmapsto\:\dfrac{dy}{y} = \dfrac{x}{ \sqrt{ {x}^{2} – 4 }}dx[/tex]

   On integrating both sides, we get

   [tex]\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dy}{y} =\displaystyle\int\tt \dfrac{x}{ \sqrt{ {x}^{2} – 4 }}dx[/tex]

   We know,

   [tex]\boxed{ \sf{ \: \displaystyle\int\tt \frac{1}{x}dx = logx + c}}[/tex]

   So, using this

   [tex]\rm :\longmapsto\:logy =\dfrac{1}{2} \displaystyle\int\tt \dfrac{2x}{ \sqrt{ {x}^{2} – 4 }}dx[/tex]

   We know,

   [tex]\boxed{ \sf{ \: \displaystyle\int\tt \frac{f'(x)}{ \sqrt{f(x)} } \: dx = 2 \sqrt{f(x)} + c}} [/tex]

   and

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} – 4) = 2x[/tex]

   So,

   [tex]\rm :\longmapsto\:logy =\dfrac{1}{2} \times 2 \sqrt{ {x}^{2} – 4 } + c[/tex]

   [tex]\rm :\longmapsto\:logy = \sqrt{ {x}^{2} – 4 } + c[/tex]

   Let’s solve one more example of same type :-

   Solve the differential equation :-

   [tex]\rm :\longmapsto\:x \sqrt{1 + {y}^{2} } dx = y \sqrt{1 + {x}^{2} }dx[/tex]

   Solution :-

   By using the method of variable separation, we get

   [tex]\rm :\longmapsto\:\dfrac{x}{ \sqrt{1 + {x}^{2} } } dx = \dfrac{y}{ \sqrt{1 + {y}^{2} } } dy[/tex]

   On integrating both sides, we get

   [tex]\rm :\longmapsto\:\displaystyle\int\tt \dfrac{x}{ \sqrt{1 + {x}^{2} } } dx =\displaystyle\int\tt \dfrac{y}{ \sqrt{1 + {y}^{2} } } dy[/tex]

   Now, multiply by 2 on both sides,

   [tex]\rm :\longmapsto\:\displaystyle\int\tt \dfrac{2x}{ \sqrt{1 + {x}^{2} } } dx =\displaystyle\int\tt \dfrac{2y}{ \sqrt{1 + {y}^{2} } } dy[/tex]

   We know,

   [tex]\boxed{ \sf{ \: \displaystyle\int\tt \frac{f'(x)}{ \sqrt{f(x)} } \: dx = 2 \sqrt{f(x)} + c}} [/tex]

   and

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} + 1) = 2x[/tex]

   So, using this, we have

   [tex]\rm :\longmapsto\:2 \sqrt{ {x}^{2} + 1 } = 2 \sqrt{ {y}^{2} + 1} + 2c[/tex]

   [tex]\rm :\longmapsto\:\sqrt{ {x}^{2} + 1 } = \sqrt{ {y}^{2} + 1} + c[/tex]

   Let’s solve one more problem now!!

   Solve the differential equation :-

   [tex]\rm :\longmapsto\:tany \: {sec}^{2}x \: dx \: = \: tanx \: {sec}^{2} y \: dy[/tex]

   Solution :-

   By method of variable separation, we have

   [tex]\rm :\longmapsto\:\dfrac{ {sec}^{2}x }{tanx}dx = \dfrac{ {sec}^{2} y}{tany} dy[/tex]

   On integrating both sides, we get

   [tex]\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {sec}^{2}x }{tanx}dx =\displaystyle\int\tt \dfrac{ {sec}^{2} y}{tany} dy[/tex]

   We know,

   [tex]\boxed{ \sf{ \: \dfrac{f'(x)}{f(x)}dx = logf(x) + c}}[/tex]

   and

   [tex]\boxed{ \sf{ \: \dfrac{d}{dx}tanx = {sec}^{2} x}}[/tex]

   So, using this we get

   [tex]\rm :\longmapsto\:log \: tanx =log \: tany + logc[/tex]

   [tex]\rm :\longmapsto\:log \: tanx =log( \: c \: tany )[/tex]

   [tex]\bf\implies \:tanx = c \: tany[/tex]

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2. Answer:

   -4xy/(x^2-4)^2

   Step-by-step explanation:

   dy/dx=xy/root x^2-4

   1/y. dy/dx= x/root x^2-4

   Solve it you will get your answer

   if I do any mistake then please forgive me becoz I’m in 11th now..

   Reply