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Ley $y=ƒ(x)$ be the solution of the differential equation $y'=\frac{3y^2+x}{4y^2+5}$ where $y(\frac{15}{4})=0$ then which of the following are correct

(A) $y'\ge\frac{3}{4} \forall x\ge \frac{15}{4}$

(B) $\int\limits_{\frac{{15}}{4}}^{\frac{{27}}{4}} {f\left( x \right)dx} \ge \frac{{27}}{8}$

(C) $\int\limits_{\frac{{15}}{4}}^{\frac{{27}}{4}} {f\left( x \right)dx} < \frac{{27}}{8}$

(D) $f'(x)\ge 0 \forall x\ge 0$

My approach is as follow $y' = \frac{{3{y^2} + x}}{{4{y^2} + 5}} \Rightarrow y' = \frac{3}{4}\left( {\frac{{4{y^2} + \frac{{4x}}{3}}}{{4{y^2} + 5}}} \right) \Rightarrow y' = \frac{3}{4}\left( {\frac{{4{y^2} + 5 + \frac{{4x}}{3} - 5}}{{4{y^2} + 5}}} \right)$

$ \Rightarrow y' = \frac{3}{4}\left( {\frac{{4{y^2} + 5}}{{4{y^2} + 5}}} \right) + \frac{3}{4}\left( {\frac{{\frac{{4x}}{3} - 5}}{{4{y^2} + 5}}} \right) \Rightarrow y' = \frac{3}{4} + \frac{1}{4}\left( {\frac{{4x - 15}}{{4{y^2} + 5}}} \right)$

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There is no need to solve the ODE to decide which of the alternatives are correct. It's clear from the expression for $y'$ that (D) is correct, and your last equation shows that (A) is also correct. And from (A) it follows that $$ f(x)=f(15/4)+\int_{15/4}^xf'(t)\,dt\geq 0+\frac{3}{4}\int_{15/4}^xdt=\frac{3}{4}\left(x-\frac{15}{4}\right) $$ for $x\geq\frac{15}{4}$, hence $$ \int_{15/4}^{27/4}f(x)\,dx\geq \frac{3}{4}\int_{15/4}^{27/4}\left(x-\frac{15}{4}\right)dx=\frac{27}{8}. $$ Therefore, (B) is correct and (C) is incorrect.

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