Question

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



​

Answer 1

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..

Answer 2

$\large\underline{\sf{Solution-}}$

Given Differential equation is,

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{xy}{ \sqrt{ {x}^{2} - 4 }}$

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

The given Differential equation can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x}{ \sqrt{ {x}^{2} - 4 }} \times y$

$\rm :\longmapsto\:\dfrac{dy}{y} = \dfrac{x}{ \sqrt{ {x}^{2} - 4 }}dx$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dy}{y} =\displaystyle\int\tt \dfrac{x}{ \sqrt{ {x}^{2} - 4 }}dx$

We know,

$\boxed{ \sf{ \: \displaystyle\int\tt \frac{1}{x}dx = logx + c}}$

So, using this

$\rm :\longmapsto\:logy =\dfrac{1}{2} \displaystyle\int\tt \dfrac{2x}{ \sqrt{ {x}^{2} - 4 }}dx$

We know,

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

and

$\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} - 4) = 2x$

So,

$\rm :\longmapsto\:logy =\dfrac{1}{2} \times 2 \sqrt{ {x}^{2} - 4 } + c$

$\rm :\longmapsto\:logy = \sqrt{ {x}^{2} - 4 } + c$

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

Solve the differential equation :-

$\rm :\longmapsto\:x \sqrt{1 + {y}^{2} } dx = y \sqrt{1 + {x}^{2} }dx$

Solution :-

By using the method of variable separation, we get

$\rm :\longmapsto\:\dfrac{x}{ \sqrt{1 + {x}^{2} } } dx = \dfrac{y}{ \sqrt{1 + {y}^{2} } } dy$

On integrating both sides, we get

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

Now, multiply by 2 on both sides,

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

We know,

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

and

$\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} + 1) = 2x$

So, using this, we have

$\rm :\longmapsto\:2 \sqrt{ {x}^{2} + 1 } = 2 \sqrt{ {y}^{2} + 1} + 2c$

$\rm :\longmapsto\:\sqrt{ {x}^{2} + 1 } = \sqrt{ {y}^{2} + 1} + c$

Let's solve one more problem now!!

Solve the differential equation :-

$\rm :\longmapsto\:tany \: {sec}^{2}x \: dx \: = \: tanx \: {sec}^{2} y \: dy$

Solution :-

By method of variable separation, we have

$\rm :\longmapsto\:\dfrac{ {sec}^{2}x }{tanx}dx = \dfrac{ {sec}^{2} y}{tany} dy$

On integrating both sides, we get

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

We know,

$\boxed{ \sf{ \: \dfrac{f'(x)}{f(x)}dx = logf(x) + c}}$

and

$\boxed{ \sf{ \: \dfrac{d}{dx}tanx = {sec}^{2} x}}$

So, using this we get

$\rm :\longmapsto\:log \: tanx =log \: tany + logc$

$\rm :\longmapsto\:log \: tanx =log( \: c \: tany )$

$\bf\implies \:tanx = c \: tany$