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How to solve dy/dx = x2 + 2xy + y2 while y(0) = 0? |
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Answer» \(\frac{dy}{dx}=x^2+2xy+y^2 = (x + y)^2\) Let x + y = t Then 1 + \(\frac{dy}{dx} = \frac{dt}{dx}\) \(\therefore\frac{dt}{dx}=1+t^2\) (\(\because\frac{dt}{dx}=(x+y)^2=t^2\)) ⇒ \(\frac{dt}{1+t^2}=dx\) ⇒ \(\int\frac{dt}{1+t^2}=\int dx\) ⇒ tan-1 t = x + c ⇒ tan-1(x + y) = x + c-----(1) (\(\because t=x+c\)) \(\because\) y(0) = 0 \(\therefore \) tan-1(0 + 0) = 0 ⇒ C = tan-1(x + y) = x is (From (1)) ⇒ x + y = tan x ⇒ y = tan x - x which is solution of given DE. |
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