# Example 28 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

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Example 28 Important Deleted for CBSE Board 2022 Exams You are here

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Dec. 11, 2019 by Teachoo

Example 28 (Introduction) Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) ππ₯/ππ¦+π₯/(1 + π¦^2 )=tan^(β1)β‘π¦/(1 + π¦^2 ) ππ₯/ππ¦=tan^(β1)β‘π¦/(1 + π¦^2 ) β π₯/(1 + π¦^2 ) ππ₯/ππ¦=(tan^(β1)β‘π¦ β π₯)/(1 + π¦^2 ) ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦ β π₯γ The variables cannot be separated. So variable separable method is not possible Now, ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦ β π₯γ Put F(x, y) = ππ¦/ππ₯ F(x, y) = (1 + π¦^2)/(tan^(β1)β‘π¦βπ₯) F(πx, πy) = (1 + π^2 π¦^2)/(tan^(β1)β‘ππ¦βππ₯)β πΒ° F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method ππ¦/ππ₯ = ((1 + π¦^2))/(tan^(β1) y β x) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We need to find ππ₯/ππ¦ ππ₯/ππ¦ = (tan^(β1)β‘π¦ β π₯)/(1 + π¦^2 ) ππ₯/ππ¦ = tan^(β1)β‘π¦/(1 + π¦^2 ) β π₯/(1 + π¦^2 ) ππ₯/ππ¦ + π₯/(1 + π¦^2 ) β (tan^(β1)β‘π¦ )/(1 + π¦^2 ) Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 Thus, we solve question by integrating factor method taking π π/π π Example 28 Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) ππ₯/ππ¦+π₯/(1 + π¦^2 )=tan^(β1)β‘π¦/(1 + π¦^2 ) Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = 1/(1 + π¦^2 ) & Q1 = (tan^(β1)β‘π¦ )/(1 + π¦^2 ) Now, IF = π^β«1βγπ_1 ππ¦γ IF = π^β«1βγ1/(1 + π¦^2 ) ππ¦γ IF = π^tan^(β1)β‘π¦ Solution is x (IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆγ xπ^tan^(β1)β‘π¦ = β«1βγtan^(β1)β‘π¦/(1 + π¦^2 )Γπ^tan^(β1)β‘π¦ ππ¦γ + C Let I = β«1βγtan^(β1)β‘π¦/(1 + π¦^2 )Γπ^tan^(β1)β‘π¦ ππ¦γ Putting tan^(β1)β‘γπ¦ γ= t 1/(1 + π¦^2 ) dy = dt Putting values of t & dt in I I = β«1βγπ‘π^π‘ ππ‘γ I = t.β«1βγπ^π‘ ππ‘ββ«1β[1β«1βγπ^π‘ ππ‘γ] ππ‘γ I = tπ^π‘ β β«1βγπ^π‘ ππ‘γ I = tπ^π‘ β π^π‘ I = π^π‘ (t β 1) Putting value of t = tan^(β1)β‘π¦ I = π^(tan^(β1)β‘π¦ ) (tan^(β1) π¦β1) Putting value of I in (1) γπ₯πγ^(tan^(β1)β‘π¦ ) = πΌ + πΆ Integrating by parts with β«1βγπ(π‘) π(π‘) ππ‘=π(π‘) β«1βγπ(π‘) ππ‘ ββ«1βγ[π^β² (π‘) β«1βγπ(π‘) ππ‘] ππ‘γγγγ Take f (t) = t & g (t) = π^"t" Divide by π^(γπ‘ππγ^(β1) π¦ ) γπ₯πγ^(tan^(β1) π¦ )/π^tan^(β1)β‘γπ¦ γ = (π^(tan^(β1) π¦) (tan^(β1)β‘π¦ β 1))/π^tan^(β1)β‘γπ¦ γ + π/π^tan^(β1)β‘γπ¦ γ π = (γπππγ^(βπ)β‘πβπ) + cπ^(βγπππγ^(βπ) π ) Which is the required general solution