Ex 9.5

Chapter 9 Class 12 Differential Equations
Serial order wise

### 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 + (π/(π πππβ‘π ))π=π/π^π Step 2: Find P and Q Comparing (1) with ππ¦/ππ₯ + Py = Q P = π/(π πππβ‘π ) & Q = π/ππ Step 3: Find Integration factor, I.F IF = e^β«1βγπ ππ₯γ IF = π^β«1βγπ/(π π₯π¨π β‘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π/ππ . log x. dx + C Let I = 2 β«1βπππβ‘γπ π^(βπ) ππγ Solving I I = 2 β«1βπππβ‘γπ π^(βπ) ππγ I = 2["log x. " β«1βγπ^(βπ) ππββ«1β π/π [β«1βγ π^(βπ) 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] = 2[(β1)/π₯ " .log x β " 1/π₯] = (βπ)/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.