1. Chapter 9 Class 12 Differential Equations
2. Serial order wise

Transcript

Example 16 Show that the differential equation π₯βπππ (π¦/π₯)=π¦ πππ (π¦/π₯)+π₯ is homogeneous and solve it. Step 1: Find ππ¦/ππ₯ π₯ πππ (π¦/π₯) ππ¦/ππ₯=π¦ cosβ‘(π¦/π₯)+π₯ ππ¦/ππ₯=(π¦ cosβ‘ (π¦/π₯) + π₯)/(π₯ cosβ‘(π¦/π₯) ) Step 2: Put F(π₯ ,π¦)=ππ¦/ππ₯ & find F(ππ₯ ,ππ¦) F(π₯ ,π¦)=(π¦ cosβ‘ (π¦/π₯) + π₯)/(π₯ cosβ‘(π¦/π₯) ) Finding F(ππ₯ ,ππ¦) F(ππ₯ ,ππ¦)=((ππ¦)πππ (ππ¦/ππ₯) + ππ₯)/((ππ¦) . cosβ‘(ππ¦/ππ₯) ) =(ππ¦ πππ (π¦/π₯) + ππ₯)/(ππ¦ cosβ‘(π¦/π₯) ) =π(π¦ πππ (π¦/π₯) + π₯)/(π π₯ cosβ‘(π¦/π₯) ) =(π¦ πππ (π¦/π₯) + π₯)/( π₯ cosβ‘(π¦/π₯) ) = F (π₯ , π¦) So , F(ππ₯ ,ππ¦)= F(π₯ , π¦) = πΒ° F(π₯ , π¦) Thus , F(π₯ , π¦) is a homogeneous function of degree zero. Therefore, the given differential equation is homogeneous differential equation Step 3: Solving ππ¦/ππ₯ by Putting π¦=π£π₯ ππ¦/ππ₯=(π¦ πππ (π¦/π₯) + π₯)/(π₯ cosβ‘(π¦/π₯) ) Put π¦=π£π₯ So, ππ¦/ππ₯=π(π£π₯) =ππ£/ππ₯ . π₯+π£ ππ₯/ππ₯ =ππ£/ππ₯ π₯+π£ Putting values of ππ¦/ππ₯ and y = vx in (1) i.e. ππ¦/ππ₯ = (π¦ πππ (π¦/π₯)+π₯)/(π₯ cosβ‘(π¦/π₯) ) ππ£/ππ₯ π₯+π£=((π£π₯) πππ (π£π₯/π₯) + π₯)/(π₯ cosβ‘(π£π₯/π₯) ) ππ£/ππ₯ π₯+π£=(π£π₯ πππ (π£) + π₯)/(π₯ cosβ‘π£ ) ππ£/ππ₯ π₯+π£=π₯(π£ cosβ‘γπ£ +1γ )/(π₯ cosβ‘π£ ) ππ£/ππ₯ π₯+π£=(π£ cosβ‘γπ£ +1γ)/cosβ‘π£ ππ£/ππ₯ π₯=(π£ cosβ‘γπ£ + 1γ)/cosβ‘π£ βπ£ ππ£/ππ₯ π₯=(π£ cosβ‘γπ£ + 1γ βπ£ cosβ‘π£)/cosβ‘π£ ππ£/ππ₯ π₯= 1/cosβ‘π£ cosβ‘π£ ππ£=ππ₯/π₯ Integrating Both Sides β«1βcosβ‘γπ£ ππ£=β«1βππ₯/π₯γ sinβ‘γπ£=logβ‘|π₯|+π1γ Putting π£=π¦/π₯ & t π1=logβ‘π π ππ π¦/π₯=logβ‘γ|π₯|+logβ‘|π| γ πππ π/π=πππβ‘|ππ|