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

Transcript

Example 28 Solve the differential equation ππ₯/ππ¦+π₯/(1+π¦^2 )=tan^(β1)β‘π¦/(1+π¦^2 ) ππ₯/ππ¦+π₯/(1 + π¦^2 )=tan^(β1)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^(β1)β‘π¦ )/(1 + π¦^2 ) Now, IF = π^β«1β"p1dy" IF = π^β«1βγ1/(1 + π¦^2 ) ππ¦γ IF = π^tan^(β1)β‘π¦ Solution is x (IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆγ xπ^tan^(β1)β‘π¦ = β«1β(tan^(β1)β‘γπ¦ π^tan^(β1)β‘π¦ γ )/(1 + π¦^2 ) dy + C Let I = β«1β(tan^(β1)β‘γπ¦ π^tan^(β1)β‘π¦ γ )/(1 + π¦^2 ) dy Putting tan^(β1)β‘γπ¦ γ= t 1/(1 + π¦^2 ) dy = dt Putting values of t & dt in I I = β«1βγπ‘π^π‘ ππ‘γ Integrating by parts with β«1βγπ(π‘) π(π‘) ππ‘=π(π‘) β«1βγπ(π‘) ππ‘ ββ«1βγ[π^β² (π‘) β«1βγπ(π‘) ππ‘] ππ‘γγγγ Take f (t) = t & g (t) = π^"t" I = t.β«1βγπ^π‘ ππ‘ββ«1β[1β«1βγπ^π‘ ππ‘γ] ππ‘γ I = tπ^π‘ β β«1βγπ^π‘ ππ‘γ I = tπ^π‘ β π^π‘ I = π^π‘ (t β 1) Putting value of t = tan^(β1)β‘π¦ I = π^(tan^(β1)β‘π¦ ) (tan^(β1) π¦β1) Putting value of I in (1) γπ₯πγ^(tan^(β1)β‘π¦ ) = πΌ + πΆ γπ₯πγ^(tan^(β1) π¦ )= π^(tan^(β1) π¦ ) (tan^(β1)β‘π¦β1) + c Divide by π^(γπ‘ππγ^(β1) π¦ ) γπ₯πγ^(tan^(β1) π¦ )/π^tan^(β1)β‘γπ¦ γ = (π^(tan^(β1) π¦) (tan^(β1)β‘π¦ β 1))/π^tan^(β1)β‘γπ¦ γ + π/π^tan^(β1)β‘γπ¦ γ π = (γπππγ^(βπ)β‘πβπ) + cπ^(βγπππγ^(βπ) π ) Which is the required general solution

Examples