Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \frac{d{y}_1}{y_1}+\frac{d{y}_2}{y_2}+\frac{d{y}_3}{y_3}=\left({\sin }^{2}x+{\cos }^{2}x+\frac{2-x^3}{x^3}\right)dx\\ & \ln \left(y_1{y}_2{y}_3\right)=\frac{-1}{x^2}+C\\ & \ln \left(y_1(x)y_2(x)y_3(x)\right)=\frac{-1}{x^2}+C\\ & \ln \left(5\cdot \frac{1}{3}\cdot \frac{3}{5e}\right)=\frac{-1}{x^2}+C\\ & \therefore C=0\end{array}$ $\begin{array}{rl}& y_1(x)y_2(x)y_3(x)=e^{\frac{-1}{x^2}}\\ & \underset{x\to 0}{\lim }\frac{e^{\frac{-1}{x^2}}+2x}{e^{3x}\sin x}\\ & \underset{x\to 0}{\lim }\frac{1}{e^{3x+\frac{1}{x^2}}\sin x}+\underset{x\to 0}{\lim }\frac{2x}{e^{3x}\sin x}\end{array}$ ${\lim }_{x\to 0}\frac{1}{e^{3x}\cdot \frac{\sin x}{x}\cdot \frac{e^{\frac{1}{x^2}}}{\frac{1}{x}}}+2$ $=0+2\left\{{\lim }_{\mathrm{y}\to \infty }\frac{{\mathrm{e}}^{{\mathrm{y}}^2}}{\mathrm{y}}=\frac{{\mathrm{e}}^{{\mathrm{y}}^2}\cdot 2\mathrm{y}}{1}=\infty \right\}$ $=2$