# Ex 9.6, 17 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

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Chapter 9 Class 12 Differential Equations (Term 2)

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Last updated at Dec. 11, 2019 by Teachoo

Ex 9.6, 17 Find the equation of a curve passing through the point(0 , 2) given that the sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 We know that Slope of tangent to curve at (x, y) = ππ¦/ππ₯ Given that sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 Therefore, |ππ¦/ππ₯| + 5 = x + y |ππ¦/ππ₯| = x + y β 5 ππ¦/ππ₯ = Β± (x + y β 5) So, we will take both positive sign and negative sign and then solve it Taking (+) ve sign ππ¦/ππ₯ = x + y β 5 ππ¦/ππ₯ β y = x β 5 Equation is of the form ππ¦/ππ₯+ππ¦=π where P = β1 & Q = x β 5 IF = e^β«1βπππ₯ IF = e^(β«1βγ(β1)γ ππ₯) IF = π^(βπ₯) Taking (β) ve sign ππ¦/ππ₯ = βx β y + 5 ππ¦/ππ₯ + y = βx + 5 Equation is of the form ππ¦/ππ₯+ππ¦=π where P = 1 & Q = βx + 5 IF = e^β«1βπππ₯ IF = e^β«1β1ππ₯ IF = e^π₯ Solution is y(IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yeβx = β«1βγ(π₯β5) π^(βπ₯) ππ₯+πγ yeβx = (x β 5) β«1βγπ^(βπ₯) ππ₯γ ββ«1βγ[1β«1βγπ^(βπ₯) ππ₯γ]ππ₯+πγ yeβx = β(x β 5)π^(βπ₯)ββ«1βγγβπγ^(βπ₯) ππ₯γ + c yeβx = β(x β 5)π^(βπ₯) + β«1βγπ^(βπ₯) ππ₯γ + c Solution is y(IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yex = β«1βγ(5βπ₯) π^π₯ ππ₯+πγ yex = (5 β x) β«1βγπ^π₯ ππ₯γ β β«1β[π/ππ₯(5βπ₯)β«1βγπ^π₯ ππ₯γ]ππ₯ yex = (5 β x) π^π₯β β«1βγ(β1)γ π^π₯ ππ₯ yex = (5 β x) π^π₯ + β«1βγπ^π₯ ππ₯γ Integrating by parts with β«1ββ(π(π₯) π(π₯) ππ₯) =π(π₯) β«1βγπ(π₯) ππ₯ γββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγ Take f (x) = x β 5 & g (x) = π^(βπ₯) Integrating by parts with β«1ββ(π(π₯) π(π₯) ππ₯) =π(π₯) β«1βγπ(π₯) ππ₯ γββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγ Take f (x) = 5 β x & g (x) = π^π₯ yeβx =(5βπ₯)π^(βπ₯) β π^(βπ₯)+π Dividing both sides by eβx y = (5 βx) β1 + cex y = 4 β x + cex Since the curve passes through the point (0, 2) Put x = 0 & y = 2 2 = 4 β 0 + ceΒ° 2 = 4 + C C = β2 β΄ Equation of curve is y = 4 β x β 2ex yex = (5 β x) π^π₯+ π^π₯+π Dividing both sides by ex y = (5 β x) + 1 + ceβx y = 6 β x + ceβx Since curve passes through the point (0, 2) Put x = 0 & y = 2 2 = 6 β 0 + ceΒ° 2 = 6 + C C = β4 β΄ Equation of curve is y = 6 β x β 4eβx