Differential Equations Question 236
Question: What is the order of the differential equation $ \frac{dx}{dy}+\int{ydx=x^{3}} $ -
Options:
A) 1
B) 2
C) 3
D) Cannot be determined
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Answer:
Correct Answer: B
Solution:
[b] $ \frac{dx}{dy}+\int{y.dx=x^{3}\Rightarrow \int{y.dx=x^{3}-\frac{dx}{dy}}} $
$ \Rightarrow 1+\frac{dy}{dx}( \int{y.dx} )=x^{3}.\frac{dy}{dx} $ Differentiate both sides w.e.t.x
$ \Rightarrow 0+\frac{dy}{dx}(y)+( \int{y.dx} )( \frac{d^{2}y}{dx^{2}} )=x^{3}.\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}(2x^{2}) $
$ \Rightarrow y.\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}[ x^{3}-\frac{dx}{dy} ]=x^{3}.\frac{d^{2}y}{dx^{2}}+2x^{2}\frac{dy}{dx} $
$ \Rightarrow y\frac{dy}{dx}+x^{3}\frac{d^{2}y}{dx^{2}}-( \frac{dx}{dy} )( \frac{d^{2}y}{dx^{2}} )=x^{3}\frac{d^{2}y}{dx^{2}}+2x^{2}\frac{dy}{dx} $
$ \Rightarrow y\frac{dy}{dx}-\frac{dx}{dy}.\frac{d^{2}y}{dx^{2}}=2x^{2}.\frac{dy}{dx} $
Multiplying both side by $ \frac{dy}{dx} $
$ y{{( \frac{dy}{dx} )}^{2}}-\frac{d^{2}y}{dx^{2}}=2x^{2}{{( \frac{dy}{dx} )}^{2}} $
$ \Rightarrow \frac{d^{2}y}{dx^{2}}+(2x^{2}-y){{( \frac{dy}{dx} )}^{2}}=0 $ Order = 2, degree = 1.
BETA
