MCQ
The solution of the differential equation $\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}$is
- ✓$4xy = {x^4} + c$
- B$xy = {x^4} + c$
- C$\frac{1}{4}xy = {x^4} + c$
- D$xy = 4{x^4} + c$
$\frac{{dy}}{{dx}} + Py = Q$. So, $I.F.$= ${e^{\int_{}^{} {\frac{1}{x}dx} }} = {e^{\log x}} = x$
Hence required solution $xy = \int_{}^{} {x.{x^2}dx + c} $
==> $xy = \frac{{{x^4}}}{4} + c$ ==> $4xy = {x^4} + c$.
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$(A)$ $y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8 \sqrt{2}}$
$(B)$ $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^2}{18}$
$(C)$ $y\left(\frac{\pi}{3}\right)=\frac{\pi^2}{9}$
$(D)$ $y ^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^2}{3 \sqrt{3}}$
The order and degree of the differential equation $\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$ are:
$(i)$ for $p \geqslant 0$ , $f(x) = 0$ has one negative root and $f(x)$ is monotonic
$(ii)$ for $-1 < p < 0$ , $f(x)$ = $0$ has one negative root and $f(x)$ is nonmonotonic
$(iii)$ for $p < 0$ , $f(x)$ = $0$ has three real and distinct roots.