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  • Let the solution curve of the differential equation <img alt="\left(4+x^{2}\right) \mathrm{d} y-2 x\left(x^{2

Let the solution curve y=y(x) of the differential equation \left(4+x^{2}\right) \mathrm{d} y-2 x\left(x^{2}+3 y+4\right) \mathrm{d} x=0 pass through the origin. Then y(2) is equal to____________.

Option: 1

12


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x=0 }\\

\mathrm{\left(x^{2}+4\right) \frac{d y}{d x}=2 x^{3}+6 x y+8 x} \\

\mathrm{\left(x^{2}+4\right) \frac{d y}{d x}-6 x y=2 x^{3}+8 x} \\

\mathrm{\frac{d y}{d x}-\frac{6 x}{x^{2}+4} y=\frac{2 x^{3}+8 x}{x^{2}+4}}

\mathrm{\text { L.I. } \frac{d y}{d x}+{P} y=\phi \quad P=\frac{-6 x}{x^{2}+4} \quad \phi=\frac{2 x^{3}+8 x}{x^{2}+4}} \\

\mathrm{\text { I.F }=e^{-\int \frac{6 x}{x^{2}+4} d \dot{x}}=e^{-3 \log _{e}\left(x^{2}+4\right)}} \\

        \mathrm{=e^{\log _{e}\left(x^{2}+4\right)^{-3}}=\frac{1}{\left(x^{2}+4\right)^{3}} }

hence, its solution is

\mathrm{y \cdot \frac{1}{\left(x^{2}+4\right)^{3}}=\int \frac{2 x^{3}+8 x}{\left(x^{2}+4\right)^{3}\left(x^{2}+4\right)} d x }\\

\mathrm{\frac{y}{\left(x^{2}+4\right)^{3}}=\int \frac{2 x\left(x^{2}+4\right)}{\left(x^{2}+4\right)^{3}\left(x^{2}+4\right)} d x \qquad\text { let } x^{2}+4=t\: so\: 2xdx= dt} \\

\mathrm{\frac{y}{\left(x^{2}+4\right)^{3}}=\int \frac{d t}{t^{3}} \Rightarrow \frac{y}{\left(x^{2}+4\right)^{3}}=\frac{-1}{2\left(x^{2}+4\right)^{2}}+c }

Passes through origin \mathrm{\left ( 0,0 \right ) }

\mathrm{0=\frac{-1}{2 \times 18}+c} \\

\mathrm{\frac{y}{\left(x^{2}+4\right)^{3}}=\frac{-1}{2\left(x^{2}+4\right)^{2}}+\frac{1}{32} \Rightarrow y=\frac{-\left(x^{2}+4\right)}{2}+\frac{\left(x^{2}+4\right)^{3}}{32} }\\

\mathrm{y(2)=-\frac{8}{2}+\frac{8 \times 8 \times 8}{32}=12}

Hence answer is \mathrm{12}

Posted by

vishal kumar

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