A) \[y=2\]
B) \[y=2x\]
C) \[y=2x-4\]
D) \[y=2{{x}^{2}}-4\]
Correct Answer: C
Solution :
Given differential equation is \[{{\left( \frac{dy}{dx} \right)}^{2}}-x\frac{dy}{dx}+y=0\] Let \[\frac{dy}{dx}=p\] \[\therefore \] \[{{p}^{2}}-px+y=0\] On differentiating w.r.t. x, we get \[2p\frac{dp}{dx}-p-x\frac{dp}{dx}+\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dp}{dx}(2p-x)=0\] \[\left[ \because \,\,\frac{dy}{dx}=p \right]\] \[\Rightarrow \] \[\frac{dp}{dx}=0\] On integrating, we get \[p=c\] \[\Rightarrow \] \[\frac{dy}{dx}=0\] On integrating, we get \[y=cx-{{c}^{2}}\] If\[c=2\], then \[y=2x-4\]
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