Examples

Chapter 9 Class 12 Differential Equations
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Example 12 Show that the differential equation 2π¦π^(π₯/π¦) ππ₯+(π¦β2π₯π^(π₯/π¦) )ππ¦=0 is homogeneous and find its particular solution , given that, π₯=0 when π¦=1 2π¦π^(π₯/π¦) ππ₯+(π¦β2π₯π^(π₯/π¦) )ππ¦ = 0 Step 1: Find ππ₯/ππ¦ 2π¦π^(π₯/π¦) ππ₯+(π¦β2π₯π^(π₯/π¦) )ππ¦=0 2π¦π^(π₯/π¦) ππ₯=β(π¦β2π₯π^(π₯/π¦) )ππ¦ 2π¦π^(π₯/π¦) ππ₯=(2π₯π^(π₯/π¦)βπ¦)ππ¦ Since the equation is in the form π₯/π¦ , we will take ππ₯/ππ¦ Instead of ππ¦/ππ₯ ππ₯/ππ¦=((2π₯π^(π₯/π¦) β π¦))/(2π¦π^(π₯/π¦) ) Step 2: Put F(π₯ , π¦)=ππ₯/ππ¦ and find F(ππ₯ ,ππ¦) F(π₯ , π¦)= (2π₯π^(π₯/π¦) β π¦)/(2π¦π^(π₯/π¦) ) Finding F(ππ ,ππ) F(ππ₯ ,ππ¦)=(2 (ππ₯) γ πγ^(ππ₯/ππ¦ βππ¦))/(2ππ¦ γ πγ^(ππ₯/ππ¦ ) )=π(2π₯π^(π₯/π¦) β π¦)/(π . 2π¦ π^(π₯/π¦) ) =(2π₯π^(π₯/π¦) β π¦)/(2π¦ π^(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 given differential equation is homogeneous differential equation Step 3: Solving ππ₯/ππ¦ by Putting π₯=π£π¦ ππ₯/ππ¦=(2π₯ π^(π₯/π¦) β π¦)/(2π¦ π^(π₯/π¦) ) Put π₯=π£π¦ Diff. w.r.t. π¦ ππ₯/ππ¦=π/ππ¦ (π£π¦) π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 x in (1) ππ₯/ππ¦=(2π₯π^(π₯/π¦) β π¦)/(2π¦ π^(π₯" " /π¦) ) π£+π¦ ππ£/ππ¦=(2π£ π^π£ β 1)/(2γ πγ^π£ ) π¦ ππ£/ππ¦=(2π£ π^π£ β 1)/(2γ πγ^π£ )βπ£ (π¦ ππ£)/ππ¦=(2π£π^π£ β 1 β 2π£π^π£)/(2γ πγ^π£ ) π¦ ππ£/ππ¦=(β1)/(2γ πγ^π£ ) 2γ πγ^π£ ππ£=(βππ¦)/( π¦) Integrating Both Sides β«1βγ2γ πγ^π£ ππ£γ=β«1β(βππ¦)/( π¦) 2γ πγ^π£=βlogβ‘|π¦|+π Putting back π£=π₯/π¦ 2π^(π₯/π¦)=βπππ|π¦|+π 2π^(π₯/π¦)+πππ|π¦|=π β¦(2) Given that at π₯=0 , π¦=1 Putting π₯=0 and π¦=1 in (2) 2π^(0/1)βπππ|1|=π 2 Γ1+0=π π=2 Put Value of π in (2) i.e., 2π^(π₯/π¦)+πππ|π¦|=πΆ ππ^(π/π)+πππ|π|=π" " is the particular solution of given differential equation (As πππ|1| = 0)