# Ex 9.3, 10 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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Ex 9.3, 23 (MCQ)

Last updated at April 16, 2024 by Teachoo

Ex 9.3, 10 For each of the differential equations in Exercises 1 to 10, find the general solution : π^π₯ tanβ‘γπ¦ ππ₯+(1βπ^π₯ ) sec^2β‘γπ¦ ππ¦=0γ γ π^π₯ tanβ‘γπ¦ ππ₯+(1βπ^π₯ ) sec^2β‘γπ¦ ππ¦=0γ γ π^π₯ tanβ‘γπ¦ ππ₯=β(1βπ^π₯ ) sec^2β‘γπ¦ ππ¦γ γ π^π₯ tanβ‘γπ¦ ππ₯γ=(π^π₯β1) sec^2β‘γπ¦ ππ¦γ π^π/(π^π β π) dx = (πππππ π π)/πππβ‘π Integrating both sides. β«1βγπ^π₯/(π^π₯ β 1) ππ₯γ = β«1βγ(π ππ2 π¦)/tanβ‘π¦ ππ¦γ Put π^πβπ = u and put tan y = v Diff u w.r.t. x ex = ππ’/ππ₯ dx = π π/ππ Diff v w.r.t. y sec2 y = ππ£/ππ¦ dy = π π/γπ¬ππγ^πβ‘π Therefore β«1βγππ₯/π’ ππ’/ππ₯γ = β«1βπ ππ2π¦/(π£ π ππ2π¦) dv β«1βππ’/π’ = β«1βππ£/π£ log u + c1 = log v Putting back u = ex β 1 and V = tan y log |"ex β 1" | + c1 = log tan y Putting c1 = log c log |"ex β 1" |+ log c = log (tan y) log |π("ex β 1" )|= log |tanβ‘π¦ | π("ex β 1" ) = tan y tan y = c ("ex β 1" ) is the general solution