# Example 22 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

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Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 22 (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 ) π π/π π = ((π + π^π))/γπππγ^(βπ)β‘γπ β πγ The variables cannot be separated. So variable separable method is not possible Now, ππ¦/ππ₯ = ((1 + π¦^2))/tan^(β1)β‘γπ¦βπ₯γ Put F(x, y) = π π/π π F(x, y) = (1 + π¦^2)/(tan^(β1)β‘π¦βπ₯) F(πx, πy) = (1 + π^2 π¦^2)/(tan^(β1)β‘ππ¦βππ₯) F(πx, πy) 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 π π/π π+π·π=πΈ β΄ We need to find π π/π π 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 π π/π π + P1 x = Q1 Thus, we solve question by integrating factor method taking π π/π π Example 22 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 π π/π π + 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^γπππγ^(βπ)β‘π Solution is x (IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆγ xπ^γπππγ^(βπ)β‘π = β«1βγγπππγ^(βπ)β‘π/(π + π^π )Γπ^γπππγ^(βπ)β‘π π πγ + C Let I = β«1βγtan^(β1)β‘π¦/(1 + π¦^2 )Γπ^tan^(β1)β‘π¦ ππ¦γ Let γπππγ^(βπ)β‘γπ γ= t 1/(1 + π¦^2 ) dy = dt Putting values of t & dt in I I = β«1βγππ^π π πγ 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" 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^(γπππ§γ^(βπ)β‘π ) (γπππγ^(βπ) πβπ) Putting value of I in (1) γπ₯πγ^(tan^(β1)β‘π¦ ) = πΌ + πΆ γπ₯πγ^(tan^(β1) π¦ )= π^(tan^(β1) π¦ ) (tan^(β1)β‘π¦β1) + c Divide by π^(γπππγ^(βπ) 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