Solving Linear differential equations - Statement given

Chapter 9 Class 12 Differential Equations
Concept wise

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Ex 9.6, 16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (π₯ , π¦) is equal to the sum of the coordinates of the point. We know that Slope of tangent to curve at (x, y) = ππ¦/ππ₯ Given that Slope of the tangent to the curve at any point (π₯ , π¦) is equal to the sum of the coordinates of the point. Therefore, ππ¦/ππ₯ = x + y ππ¦/ππ₯ β y = x This is the form ππ¦/ππ₯+Py=Q where P = β1 & Q = x Finding Integrating factor IF = π^β«1βγπππ₯ γ IF = π^β«1βγ(β1)ππ₯ γ IF = eβx Solution is y(IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ π¦π^(βπ₯) = β«1βγπ₯π^(βπ₯) ππ₯+πγ yeβx = x β«1βγπ^(βπ₯) ππ₯βγ β«1βγ[1β«1βγπ^(βπ₯) ππ₯γ] ππ₯+πγ yeβx = βx π^(βπ₯) ββ«1βγβπ^(βπ₯) ππ₯+πγ yeβx = βx π^(βπ₯)+β«1βγπ^(βπ₯) ππ₯+πγ yeβx = βx π^(βπ₯)+(βπ^(βπ₯))/(β1) +π yeβx = βx π^(βπ₯)βπ^(βπ₯) +π Dividing both sides by eβx y = βx β 1 + π/π^(βπ₯) 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 (x) = x & g (x) = π^(βπ₯) y = βx β 1 + cex Since curve passes through origin, Putting x = 0 & y = 0 in (2) 0 = 0 β 1 + Ce0 1 = C C = 1 Put value of C in (1) y = βx β 1 + ex x + y + 1 = ex β¦(1)