A) 1, 1
B) 1, 2
C) 2, 1
D) None of these
Correct Answer: B
Solution :
The given differential equation is \[y=px+\sqrt{{{a}^{2}}{{p}^{2}}+{{b}^{2}}},p=\frac{dy}{dx}\] \[\Rightarrow \]\[y-px=\sqrt{{{a}^{2}}{{p}^{2}}+{{b}^{2}}}\] On squaring both sides, we get \[\Rightarrow \]\[{{y}^{2}}+{{p}^{2}}{{x}^{2}}-2pyx={{a}^{2}}{{p}^{2}}+{{b}^{2}}\] \[\Rightarrow \]\[{{y}^{2}}+{{\left( \frac{dy}{dx} \right)}^{2}}{{x}^{2}}-2\left( \frac{dy}{dx} \right)xy={{a}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}+{{b}^{2}}\] Clearly, this is the differential equation of degree and order 1.
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