If the curve


If the curve $y=y(x)$ is the solution of the differential equation

$2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0$

which passes through the point $\left(1,1-\frac{4}{3} \log _{e} 2\right)$, then the value of $y(16)$ is equal

to :

  1. $4\left(\frac{31}{3}+\frac{8}{3} \log _{e} 3\right)$

  2. $\left(\frac{31}{3}+\frac{8}{3} \log _{e} 3\right)$

  3. $4\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{e}} 3\right)$

  4. $\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{c}} 3\right)$

Correct Option: , 3


$\frac{d y}{d x}-\frac{y}{2 x}=\frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)}$

$\mathrm{IF}=\mathrm{e}^{-\int \frac{\mathrm{dx}}{2 \mathrm{~d}}}=\mathrm{e}^{-\frac{1}{2} \ln \mathrm{x}}=\frac{1}{\mathrm{x}^{1 / 2}}$

$y \cdot x^{-1 / 2}=\int \frac{x^{9 / 4} \cdot x^{-1 / 2}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} d x$

$\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x$

$x=t^{4} \Rightarrow d x=4 t^{3} d t$

$\int \frac{t^{2} \cdot 4 t^{3} d t}{\left(t^{3}+1\right)}$

$4 \int \frac{\mathrm{t}^{2}\left(\mathrm{t}^{3}+1-1\right)}{\left(\mathrm{t}^{3}+1\right)} \mathrm{dt}$

$4 \int \mathrm{t}^{2} \mathrm{dt}-4 \int \frac{\mathrm{t}^{2}}{\mathrm{t}^{3}+1} \mathrm{dt}$

$\frac{4 t^{3}}{3}-\frac{4}{3} \ln \left(t^{3}+1\right)+C$

$\mathrm{yx}^{-1 / 2}=\frac{4 \mathrm{x}^{3 / 4}}{3}-\frac{4}{3} \ln \left(\mathrm{x}^{3 / 4}+1\right)+\mathrm{C}$

$1-\frac{4}{3} \log _{\mathrm{e}} 2=\frac{4}{3}-\frac{4}{3} \log _{\mathrm{e}} 2+\mathrm{C}$

$\Rightarrow \mathrm{C}=-\frac{1}{3}$

$y=\frac{4}{3} x^{5 / 4}-\frac{4}{3} \sqrt{x} \ln \left(x^{3 / 4}+1\right)-\frac{\sqrt{x}}{3}$

$y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \ln 9-\frac{4}{3}$

$=\frac{124}{3}-\frac{32}{3} \ln 3=4\left(\frac{31}{3}-\frac{8}{3} \ln 3\right)$

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