Ex 9.5

Chapter 9 Class 12 Differential Equations
Serial order wise

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### Transcript

Ex 9.5, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : π₯ππππ₯ ππ¦/ππ₯+π¦=2/π₯ ππππ₯ Step 1: Put in form ππ¦/ππ₯ + Py = Q xlog x ππ¦/ππ₯ + y = 2/π₯ log x Dividing by x log x, ππ¦/ππ₯+π¦" Γ " 1/(π₯ logβ‘π₯ ) = 2/π₯ πππ π₯" Γ " 1/(π₯ logβ‘π₯ ) ππ¦/ππ₯ + (1/(π₯ logβ‘π₯ ))π¦=2/π₯^2 Step 2: Find P and Q ...(1) Comparing (1) with ππ¦/ππ₯ + Py = Q P = 1/(π₯ logβ‘π₯ ) & Q = 2/π₯2 Step 3: Find Integration factor, I.F IF = e^β«1βγπ ππ₯γ IF = e^β«1βγ1/(π₯ logβ‘π₯ ) ππ₯γ Let t = log x dt = 1/π₯ dx dx = x dt So, IF = e^β«1βγ1/(π₯ π‘) Γ π₯ππ‘γ IF = e^β«1βγ1/π‘ ππ‘γ IF = e^logβ‘γ|π‘|γ IF = |π‘| Putting back t = log x IF = |log x| IF = log x Step 4: Solution of the equation y Γ I.F = β«1βγπΓπΌ.πΉ. ππ₯+πΆγ Putting values, y Γ log x = β«1β2/π₯2 . log x. dx + C (As log x is always positive) ...(2) Let I = 2 β«1βlogβ‘γπ₯ π₯^(β2) ππ₯γ Solving I I = 2 β«1βlogβ‘γπ₯ π₯^(β2) ππ₯γ I = 2["log x. " β«1βγπ₯^(β2) ππ₯ββ«1β 1/π₯ [β«1βγ π₯^(β2) ππ₯γ] γ ππ₯" " ] I = 2 ["log x . " π₯^(β1)/((β1)) " β " β«1βγ 1/π₯γ " . " ((π₯^(β1)))/((β1)) ".dx " ] = 2["β log x. " 1/π₯ " + " β«1βγ1/π₯^2 .ππ₯γ] Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = log x & g (x) = π₯^(β2) = 2[(β1)/π₯ " .log x β " 1/π₯] = (β2)/π₯ (1 + log x) Putting value of I in (2) y log x = I + C y. log x = (βπ)/π (1 + log x) + C Which is the general solution of the given equation.