1. Class 12
2. Important Question for exams Class 12

Transcript

Misc 11 Find a particular solution of the differential equation (π₯βπ¦)(ππ₯+ππ¦)=ππ₯βππ¦ , given that π¦=β1 , when π₯=0 (π»πππ‘:ππ’π‘ π₯βπ¦=π‘) (π₯βπ¦)(ππ₯+ππ¦)=ππ₯βππ¦ π₯ππ₯ + π₯ππ¦ β y dx β y dy = dx β dy x dx β y dx β dx = β xdy + y dy β dy (x β y β 1) dx = (βx + y β 1)dy ππ¦/ππ₯ = (π₯ β π¦ β 1)/(βπ₯ + π¦ β 1) ππ¦/ππ₯ = ((π₯ β π¦ β 1))/(β(π₯ βπ¦ + 1)) Put x β y = t Diff w.r.t.x 1 β ππ¦/ππ₯ = ππ‘/ππ₯ Putting value of t & dt in (1) 1 β ππ‘/ππ₯ = β ((π‘ β 1))/(π‘ + 1) ππ‘/ππ₯ = 1 + ((π‘ β 1))/(π‘ + 1) ππ‘/ππ₯ = (π‘ + 1 + π‘ β 1)/(π‘ + 1) ππ‘/ππ₯ = 2π‘/(π‘ β 1) (π‘ β 1)/2π‘ dt = ππ₯ Integrating both sides β«1βγ(π‘ + 1 )/2π‘ ππ‘γ = β«1βππ₯ β«1βγ(π‘/2π‘+1/2π‘) ππ‘γ = x + c β«1β(1/2+1/2π‘)ππ‘ = x + c π‘/2 + 1/2 log |π‘| = x + c Putting value of t = x β y (π₯ β π¦)/2 + 1/2 log |π₯βπ¦| = x + C Given y = β1 when x = 0 Put x = 0 & y = β1 in (2) (0 + 1)/2 + 1/2 log 1 = 0 + C 1/2 = C C = 1/2 Putting value in (2) (π₯ β π¦)/2+1/2 logβ‘γ|π₯ β π¦|=π₯+1/2γ (π₯ β π¦)/2+1/2 logβ‘γ|π₯ β π¦|=(2π₯ + 1)/2γ x β y + log |π₯βπ¦|=2π₯+1 log |π₯βπ¦| = 2x + 1 β x + y log |πβπ| = x + y + 1

Class 12
Important Question for exams Class 12