Question

Solution of $\frac{d^{2}y}{d{x}^2}$= $\log x$ is:

• A. $y=\frac{1}{2}{x}^2\log x-\frac{3}{4}{x}^2+c_{1}x+c_2$
• B. $y=\frac{1}{2}{x}^2\log x+\frac{3}{4}{x}^2+c_{1}x+c_2$
• C. $y=\frac{-1}{2}{x}^2\log x-\frac{3}{4}{x}^2+c_{1}x+c_2$
• D. $y=3x^2+4x+c_1$

Answer 1

The correct option is A $y=\frac{1}{2}{x}^2\log x-\frac{3}{4}{x}^2+c_{1}x+c_2$

$\frac{d^{2}y}{d{x}^2}=\log x$

$\Rightarrow \frac{d}{dx}{y}_1=\log x$

$\Rightarrow \int d{y}_1=\int \log xdx+c_1$

$\int \log dx=x\log x-x+c_1$

$\Rightarrow y_1=x\log x-x+c_1$

$\int dy=\int (x\log x-x+c_1)dx+c_2$

$y=\int x\log xdx-\frac{x^2}{2}+c_{1}x+c_2$

$\int x\log xdx=x(x\log x-x)-\int (x\log x-x)dx$

$\to 2\int x\log xdx=x^2\log x-x^2+x^2{/}_2=x^2\log x^{-}{x}^2{/}_2$

$\Rightarrow \int x\log xdx=\frac{x^2\log x}{2}-x^2{/}_4$

$\Rightarrow y=\frac{x^2\log x}{2}-\frac{3x^2}{4}+c_{1}x+c_2$