Question

Solve!!

Differential equations.


Using integrations!!

Answer 1

The answer is explained in the attachment.

Hope it helps!

Answer 2

Hope u like my process
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The differential Equations in the form of

$\frac{dy}{dx} + py = q.....(1)$
can be solved into general solutions by following process :-

1) find out the integration factor.

=> Here integration factor refers to the value that comes after solving:-

$= > \it \: {e}^{ \int \: p \: dx}$
2) multiply Integrating factor on both sides.

3) integrate both sides to get solution.
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xi)

$= > \it \: (1 + {x}^{2} ) \frac{dy}{dx} + y = \tan ^{ - 1} (x) \\ \\ = > \bf \frac{dy}{dx} + ( \frac{1}{1 + {x}^{2} } )y = \frac{ \tan {}^{ - 1} (x) }{1 + {x}^{2} }$

Clearly, here

$= > p = \frac{1}{1 + {x}^{2} } \\ \\ \bf \underline{ so \: \: integrating \: \: factor \: \: is} \\ \\ = > {e}^{ \int \: p \: dx} = {e}^{ \int \: \frac{dx}{1 + {x}^{2} } } = {e}^{ \tan {}^{ - 1} (x) }$

Thus,

Multiplying I. F. both sides we get,
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$= > {e}^{ \tan {}^{ - 1} (x) } \frac{dy}{dx} + \frac{ {e}^{ \tan {}^{ - 1} (x) } }{(1 + {x}^{2}) } y = \frac{ \tan {}^{ - 1} (x) {e}^{ \tan {}^{ - 1} (x) } }{(1 + {x}^{2} )} \\ \\ \: \: \: \: \: \: \: \: \: \bf \: solving \: \: we \: \: get \\ \\ = > \int \: d(y {e}^{ \tan {}^{ - 1} ( x ) } ) = \int \frac{ \tan {}^{ - 1} (x) {e}^{ \tan {}^{ - } (x) } }{1 + {x}^{2} } dx \\ \\ \bf \: \: \: let \: \: \tan {}^{ - 1} (x) = z \\ \\ so.. \: \: \: \it \frac{dx}{1 + {x}^{2} } = dz \\ \\ \bf \: \: now \\ \\ = > y. {e}^{ z } = \int \: z. {e}^{z}dz \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = z \int \: {e}^{z} dz - \int \frac{dz}{dz} ( \int {e}^{z}d z)dz \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {e}^{z} (z - 1) + c \\ \\ \\ = > \it \: y = z - 1 + c {e}^{ - z} \\ \\ = > \underline{\bf \: y = \tan {}^{ - 1} (x) - 1 + c {e}^{ - \tan {}^{ - 1} (x) } } \\ \\ this \: \: is \: \: your \: \: required \: \: general \: \: solution$

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xii)

$= > x log(x) \frac{dy}{dx} + y = 2 log(x) \\ \\ = > \bf \: \frac{dy}{dx} + ( \frac{1}{x log(x) } )y = \frac{2}{x}$

Thus clearly..

$= > p = \frac{1}{x log(x) } \\ \\ \bf \: \underline{taking \: \: out \: \: integration \: \: factor} \\ \\ = > {e}^{ \int \: p\: dx} = {e}^{ \int \frac{dx}{x log(x) } } = {e}^{ \int \frac{ \frac{1}{x}dx }{ log(x) } } \\ \\ = \it \: {e}^{ log( log(x) ) } = \bf log(x) \\ \\ \bf \: now \: \: \\ \\ \underline{multipying \: \: integrating \: \: factor\: \: in \: \: both \: \: sides } \\ \\ = > \it \: log(x) \frac{dy}{dx} + \frac{1}{x} .y = \frac{2 log(x) }{x} \\ \\ = > \int \: d(y. log(x) ) = 2 \int \frac{ log(x) }{x} dx \\ \\ \bf \: let \: \: log(x) = z \\ \\ so... \frac{1}{x} dx = dx \\ \\ \bf \: \: \: \: \: \: \: \underline{now} \\ \\ = > \it y \times z = 2 \int \: zdz \\ \\ = > y \times z = 2 \times \frac{ {z}^{2} }{2} + c \\ \\ = > y = z + \frac{c}{z} \\ \\ = > \bf \: \underline{y = log(x) + \frac{c}{ log(x) } }$
This is the required general solution.
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Hope this is ur required answer

Proud to help you