Question

Hard

Solving time: 4 mins

# Find the integrating factor for the differential equation

$(3+5cosx)dxdy +y=x_{2}$

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## Text solutionVerified

We have, $(3+5cosx)dxdy +y=x_{2}$

$⇒dxdy +3+5cosxy =3+5cosxx_{2} $

So, I.F $_{.}=e_{∫pdx}$ , where $p=3+5cosx1 $

Let $I=∫3+5cosx1 dx$

Let $tan2x =t⇒sec_{2}2x ×21 dx=dt$

$⇒dx=sec22x 2dt =1+t_{2}2dt $

Also, $cosx=1+t_{2}1−t_{2} $

So, $I=∫3+5(1+t1−t )1+t2dt =∫8−2t_{2}2dt =∫4−t_{2}dt =41 g∣∣ 2−t2+t ∣∣ =g∣∣ 2−tan2x 2+tan2x ∣∣ _{41}$

So, $I.F.=e_{log∣∣2−tan2x2+tan2x∣∣_{41}}=∣∣ 2−tan2x 2+tan2x ∣∣ _{41}$

So, the correct option is $(A)$ .

$⇒dxdy +3+5cosxy =3+5cosxx_{2} $

So, I.F $_{.}=e_{∫pdx}$ , where $p=3+5cosx1 $

Let $I=∫3+5cosx1 dx$

Let $tan2x =t⇒sec_{2}2x ×21 dx=dt$

$⇒dx=sec22x 2dt =1+t_{2}2dt $

Also, $cosx=1+t_{2}1−t_{2} $

So, $I=∫3+5(1+t1−t )1+t2dt =∫8−2t_{2}2dt =∫4−t_{2}dt =41 g∣∣ 2−t2+t ∣∣ =g∣∣ 2−tan2x 2+tan2x ∣∣ _{41}$

So, $I.F.=e_{log∣∣2−tan2x2+tan2x∣∣_{41}}=∣∣ 2−tan2x 2+tan2x ∣∣ _{41}$

So, the correct option is $(A)$ .

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**LIVE**classesQuestion Text | Find the integrating factor for the differential equation $(3+5cosx)dxdy +y=x_{2}$ |

Updated On | Oct 14, 2022 |

Topic | Differential Equations |

Subject | Mathematics |

Class | Class 12 |

Answer Type | Text solution:1 Video solution: 1 |

Upvotes | 118 |

Avg. Video Duration | 3 min |