# Ex 9.2, 6 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 9.2, 6 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation : π¦=π₯ sinβ‘π₯: π₯π¦^β²=π¦+π₯β(π₯^2βπ¦^2 ) (π₯β 0 πππ π₯>π¦ ππ π₯<βπ¦ ) π¦=π₯ sinβ‘π₯ ππ¦/ππ₯=(π(π₯ sinβ‘π₯))/ππ₯ =π(π₯)/ππ₯.sinβ‘π₯+π₯.π(sinβ‘π₯ )/ππ₯ =1.sinβ‘π₯+π₯(cosβ‘π₯ ) =sinβ‘π₯+π₯cosβ‘π₯ Now, we have to verify π₯π¦^β²=π¦+π₯β(π₯^2βπ¦^2 ) L.H.S π₯π¦^β² =π₯[sin π₯+π₯πππ π₯] =π₯ π ππ π₯+π₯^2 cos π₯ R.H.S π¦+π₯β(π₯^2βπ¦^2 ) =π₯ π ππ π₯+π₯β(π₯^2β(π₯ π ππ π₯)^2 ) =π₯ π ππ π₯+π₯β(π₯^2 (1βsin^2β‘π₯ ) ) =π₯ π ππ π₯+π₯^2 β(cos^2β‘π₯ ) =π₯ π ππ π₯+π₯^2 cosβ‘π₯ Since LHS = R.H.S Hence Verified