A) \[x{{\left( y' \right)}^{2\text{ }}}=x-2yy'\]
B) \[x{{\left( y' \right)}^{2}}=2yy'-x\]
C) \[x{{\left( y' \right)}^{2}}=x+2yy'\]
D) \[xy''=y'\]
Correct Answer: C
Solution :
\[{{x}^{2}}=4b\left( y+b \right)\] ... (i) \[\Rightarrow \,\,\,\,\,\,2x=4b\left( \frac{dy}{dx} \right)\] \[\Rightarrow \,\,\,\,x=2b\frac{dy}{dx}\] \[\Rightarrow \,\,\,\,b=\frac{x}{2\left( \frac{dy}{dx} \right)}\] ?. (ii) Put b from (ii) in (i) \[\Rightarrow \,\,\,\,{{x}^{2}}=\frac{4\times x}{2\times \frac{dy}{dx}}\left( y+\frac{x}{2\left( \frac{dy}{dx} \right)} \right)\] \[\Rightarrow \,\,\,\,x\frac{dy}{dx}=\frac{\left( 2y\frac{dy}{dx}+x \right)\times 2}{2\left( \frac{dy}{dx} \right)}\] \[\Rightarrow \,\,\,\,x{{\left( \frac{dy}{dx} \right)}^{2}}=2y\frac{dy}{dx}+x\]
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