Formation of Differntial equation when general solution given
Formation of Differntial equation when general solution given
Last updated at August 5, 2026 by Teachoo
Transcript
Question 9 Form the differential equation of the family of hyperbolas having foci on š„āšš„šš and center at origin. Equation of hyperbola having foci on x-axis & center at origin (0, 0) is š„^2/š^2 āš¦^2/š^2 =1 ā“ Differentiating Both Sides w.r.t. š„ š/šš„ [š„^2/š^2 āš¦^2/š^2 ]=š(1)/šš„ 1/š^2 [2š„]ā1/š^2 [2š¦ . šš¦/šš„]=0 2š¦/š^2 . š¦ā²=2š„/š^2 Since it has two variables, we will differentiate twice š¦/š^2 š¦ā²=š„/š^2 (š¦/š„)š¦ā²=š^2/š^2 (š¦š¦^ā²)/š„ = š^2/š^2 Again differentiating both sides w.r.t. x ((š¦š¦^ā² )^ā² š„ ā (šš„/šš„)(š¦š¦^ā² ))/š„^2 =0 (š¦š¦^ā² )^ā² š„ ā (1)(š¦š¦^ā² )=šĆš^š (š¦š¦^ā² )^ā² š„ āš¦š¦^ā²=š (šš^ā² )^ā² š„ āš¦š¦^ā²=0 (Using Quotient rule and Diff. of constant is 0) (š^ā² š^ā²+ššā²ā²)š„ āš¦š¦^ā²=0 (暦^ā²ć^2+š¦š¦ā²ā²)š„ āš¦š¦^ā²=0 š„暦^ā²ć^2+š„š¦š¦^ā²ā²āš¦š¦^ā²=0 ššš^ā²ā²+šćš^ā²ć^šāšš^ā²=š (Using Product rule)