Question

Plzzzz solve this. ​

Answer 1

Answer:

dx = 1/1+ysquare -2y dy

Step-by-step explanation:

first of all

1+ysquare dx = taninverse -x dx

1+x dx = taninverse -ysquare dy

differentiating x term with respect to x & y term with respect to y,

hence ,

0 +1 dx = 1/1+y square -2y dy

dx = 1/1+y y -2y dy + c

Answer 2

Question:

Solve the differential equation:

$(1 + {y}^{2})dx = ( {tan}^{ - 1} - x)dy$

Solution:

The given equation can be written as:

$\frac{dx}{dy} + \frac{x}{1 + {y}^{2} } = \frac{ {tan}^{ - 1}y }{1 + {y}^{2} } .........(1)$

Comparing it with

$\frac{dx}{dy} + px = q$

We have:

$p = \frac{1}{1 + {y}^{2} } \: \: and \: \: q = \frac{ {tan}^{ - 1} y}{1 + {y}^{2} } \\ \\ I.F .= {e}^{\int_ \: p . \: dy } = {e}^{\int_\: \frac{1}{1 + {y}^{2} }dy } = {e}^{ {tan}^{ - 1y} }$

Multiplying (1) by ,

${e}^{tan ^{ - 1}y }$

we get:

${e}^{ {tan}^{ - 1} }. \frac{dx}{dy} + \frac{x}{1 - {y}^{2} } {e}^{ {tan}^{ - 1} y} = \frac{ {tan}^{ - 1}y }{1 + {y}^{2} } . {e}^{ {tan}^{ - 1} y} \\ \\ = > \frac{d}{dy} (x. {e}^{ { {tan}^{ - 1}y } } ) = ( \frac{ {tan}^{ - 1} y}{1 + {y}^{2} } ) {e}^{ {tan}^{ - 1} y}$

Integrating,

$x {e}^{tan ^{ - 1} y} = \int_ \: \: ( \frac{ {tan}^{ - 1} y}{1 + {y}^{2} }) {e}^{ {tan}^{ - 1} y} dy + c.........(2)$

Now

$I = \int_ \: (\frac{ {tan}^{ - 1}y }{1 + {y}^{2} } ) {e}^{ {tan}^{ - 1}y } dy$

$\boxed{put \: {tan}^{ - 1} = t }$

So that

$( \frac{1}{1 + {y}^{2} } )dy = dt \\ \\ I = \int_ \: t \: {e}^{t} \: dt \\ \\ integrating \: by \: parts \\ \\ \: \: = t {e}^{t} - \int_ \: (1) {e}^{t} dt \\ \\ \: \: = t {e}^{t} - {e}^{t} \\ \\ \: \: = {e}^{t} (t - 1) \\ \\ \: \: = {e}^{{tan}^{ - 1} y} ( {tan}^{ - 1} y - 1)$

From (2)

$x \: {e}^{ {tan}^{ - 1} y} = {e}^{ {tan}^{ - 1} y} ( {tan}^{ - 1} y - 1) + c \\ \\ x = ( {tan}^{ - 1} y - 1) + c \: {e}^{ { - tan}^{ - 1} y} ............(3)$

Which is required solution.