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00:00 - 00:59 | hello Gai To Kuchh Nahin find the general solution of the differential equation that is 1 + x square + 2 X Y DX and context react now we can write it as DY by DX DY by DX is equal to x minus 2 x y divided by 1 + x square and begin again right time in simplified form DY by DX + 2 x y divided by 1 + x square is equal to cot X divided by 1 + x square now here you see that it is the form of DY by DX + 3 Y is equal to q now where p is equal to P is equal to 2 x divided by 1 + x square and cube is equal to cot X divided by 1 + x square |

01:00 - 01:59 | now use the formula and first we find the integrating factor that is equal to integration of b.ds that is equal to first we take 1 + x square is equal to tea in p u c that 2 x divided by 1 + x square now we take now we take 1 + x square is equal to tea so we get 2x DX is equal to didi and in place of DXB right DT divided by 2 x DT / so 2x 2x cancel each other we left with 181 by it and integration of DTD LNT so its integration is Allen of t so it is equal to 3 and it is equality value is 1 + x square oneplus Aqua this is our integrating factor now y dot |

02:00 - 02:59 | white out integrating factor is equal to integration of q q value is cortex divided by 1 + x square and multiply with integrating factor integrating factor is 1 + x square and DX now 1 + x square 1 + x square cancel each other we left with white dot integrating factor is 1 + X + 1 + x square and integration of cot X is equal to log of sin x is equal to log of sin x + c this is our value of Y is equal to log sin x divided by 1 + x square and + c divided by 1 + x square so this is our answer of the following question bank you |

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