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Last updated at Dec. 11, 2019 by Teachoo
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Misc 2 For each of the exercise given below , verify that the given function (ππππππππ‘ ππ ππ₯ππππππ‘) is a solution of the corresponding differential equation . (i) π₯π¦=π π^π₯+π π^(βπ₯)+π₯^2 : π₯ (π^2 π¦)/(ππ₯^2 )+2 ππ¦/ππ₯βπ₯π¦+π₯^2β2=0 π₯π¦=ππ^π₯+π π^(βπ₯)+π₯^2 Differentiating w.r.t x (π(π₯π¦))/ππ₯=π/ππ₯ [π π^π₯+π π^(βπ₯)+π₯^2 ] ππ₯/ππ₯ y+ππ¦/ππ₯ π₯ =πγ πγ^π₯+(β1)π π^(βπ₯)+2π₯ y+y^β² x =aγ eγ^xβb e^(βx)+2x Differentiating again w.r.t x π¦β²+(π¦^β² π₯)^β² =(ππ^π₯ )^β²β(ππ^(βπ₯) )^β²+(2π₯)^β² π¦^β²+(π¦^β²β² π₯+π¦^β²Γ1)=ππ^π₯+ππ^(βπ₯)+2 π¦^β²β² π₯+2π¦^β²=ππ^π₯+ππ^(βπ₯)+2 Now, we know that π₯π¦=ππ^π₯+π π^(βπ₯)+π₯^2 π₯π¦βπ₯^2=ππ^π₯+π π^(βπ₯) ππ^π₯+π π^(βπ₯)=π₯π¦βπ₯^2 ...(1) (Given equation) ...(2) Putting (2) in (1) π¦^β²β² π₯+2π¦^β²=ππ^π+ππ^(βπ)+2 π¦^β²β² π₯+2π¦^β²=ππβπ^π+2 (π^2 π¦)/(ππ₯^2 ) π₯+2 ππ¦/ππ₯=ππβπ^π+2 (π ^π π)/(π π^π ) π+π π π/π πβππ+π^π=π β΄ The given function is a solution Misc 2 For each of the exercise given below , verify that the given function (ππππππππ‘ ππ ππ₯ππππππ‘) is a solution of the corresponding differential equation . (ii) π¦=π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ ) : (π^2 π¦)/(ππ₯^2 )β2 ππ¦/ππ₯+2π¦=0 π¦=π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ ) : Differentiating w.r.t. x π¦^β²=[π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ )]^β² π¦^β²=γ(πγ^π₯)β² (π cosβ‘γπ₯+π sinβ‘π₯ γ )+π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ )β² π¦^β²=π^π (π πππβ‘γπ+π πππβ‘π γ )+π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ ) Putting π¦=π^π₯ (π πππ β‘γπ₯+π π ππβ‘π₯ γ ) : π¦^β²=π+π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ ) π¦^β²βπ¦=π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ ) Differentiating again w.r.t x π¦^β²β²βπ¦^β²=[π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ )]^β² π¦^β²β²βπ¦^β²=γ(πγ^π₯)β²(βπ sinβ‘γπ₯+π cosβ‘π₯ γ )+π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ )β² π¦^β²β²βπ¦^β²=π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ )+π^π₯ (βπ cosβ‘γπ₯βπ siπβ‘π₯ γ ) π¦^β²β²βπ¦^β²=π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ )βπ^π₯ (π cosβ‘γπ₯+π siπβ‘π₯ γ ) Putting π¦=π^π₯ (π πππ β‘γπ₯+π π ππβ‘π₯ γ ) π¦^β²β²βπ¦^β²=π^π₯ (βπ sinβ‘γπ₯+π cosβ‘π₯ γ )βπ¦ Putting π¦^β²βπ¦=π^π₯ (βπ π ππβ‘γπ₯+π πππ β‘π₯ γ ) from (1) β¦(1) π¦^β²β²βπ¦^β²=π¦^β²βπ¦βπ¦ π¦^β²β²βπ¦^β²=π¦^β²β2π¦ π¦^β²β²βπ¦^β²βπ¦^β²+2π¦=0 π¦^β²β²β2π¦^β²+2π¦=0 (π^2 π¦)/(ππ₯^2 )β2 ππ¦/ππ₯+2π¦=0 Thus, Given Function is a solution of the Differential Equation Misc 2 For each of the exercise given below , verify that the given function (ππππππππ‘ ππ ππ₯ππππππ‘) is a solution of the corresponding differential equation . (iii) π¦=π₯ sinβ‘3π₯ : (π^2 π¦)/(ππ₯^2 )+9π¦β6 cosβ‘γ3π₯=0γ π¦=π₯ sinβ‘3π₯ Differentiating w.r.t x π¦^β²=(π₯ π ππβ‘3π₯ )^β² π¦^β²=π₯^β² sinβ‘3π₯+π₯(sinβ‘3π₯)β² π¦^β²=sinβ‘3π₯+π₯Γ3 cosβ‘3π₯ π¦^β²=sinβ‘3π₯+3π₯ cosβ‘3π₯ Differentiating again w.r.t. x π¦^β²β²=(sinβ‘3π₯ )^β²+(3π₯ cosβ‘3π₯ )^β² π¦^β²β²=3 cosβ‘3π₯+γ3(π₯)γ^β² cosβ‘3π₯+3π₯ (cosβ‘3π₯ )^β² π¦^β²β²=3 cosβ‘3π₯+3 cosβ‘3π₯+3π₯(β3 sinβ‘3π₯) π¦^β²β²=6 cosβ‘3π₯β9π₯ sinβ‘3π₯ Putting π¦=π₯ sinβ‘3π₯ π¦^β²β²=6 cosβ‘3π₯β9π¦ π¦^β²β²+9π¦β6 cosβ‘3π₯=0 Thus, Given Function is a solution of the Differential Equation Misc 2 For each of the exercise given below , verify that the given function (ππππππππ‘ ππ ππ₯ππππππ‘) is a solution of the corresponding differential equation . (iv) π₯^2=2π¦^2 logβ‘π¦ : (π₯^2+π¦^2 ) ππ¦/ππ₯βπ₯π¦=0 π₯^2=2π¦^2 logβ‘π¦ Differentiating Both sides w.r.t. x (π₯^2 )^β²=(2π¦^2 logβ‘π¦ )β² 2π₯=(2π¦^2 )^β² logβ‘π¦ +2π¦^2 (logβ‘π¦ )^β² 2π₯=2Γ2π¦π¦^β² logβ‘π¦ + 2π¦^2Γ 1/π¦ π¦β² 2π₯=4π¦π¦^β² logβ‘π¦ + 2π¦π¦β² π₯=2π¦π¦^β² logβ‘π¦ + π¦π¦β² π₯=γπ¦π¦γ^β² (2 logβ‘π¦+1) Now, from our equation π₯^2=2π¦^2 logβ‘π¦ π₯^2/(2π¦^2 )=logβ‘π¦ πππβ‘π¦=π₯^2/(2π¦^2 ) Putting value of πππβ‘π¦ in (1) π₯=γπ¦π¦γ^β² (2 πππβ‘π+1) π₯=γπ¦π¦γ^β² (2Γπ₯^2/(2π¦^2 ) " " +1) π₯=γπ¦π¦γ^β² (π₯^2/π¦^2 " " +1) π₯=γπ¦π¦γ^β² ((π₯^2 + π¦^2)/π¦^2 ) π₯=π¦^β² ((π₯^2 + π¦^2)/π¦) π₯π¦=π¦^β² (π₯^2 + π¦^2 ) π¦^β² (π₯^2 + π¦^2 )βπ₯π¦=0 Thus, Given Function is a solution of the Differential Equation
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