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For the equation $\dfrac{dy}{dx}+7x^2y=0$, if $y(0)=3/7$, then the value of $y(1)$ is

  1. $\dfrac{7}{3}e^{-7/3} \\$
  2. $\dfrac{7}{3}e^{-3/7} \\$
  3. $\dfrac{3}{7}e^{-7/3} \\$
  4. $\dfrac{3}{7}e^{-3/7}$
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Given Differential Equation is :-

$\frac{\mathrm{d} y}{\mathrm{d} x}\; +7x^{2}y = 0$

On separating $x$ and $dx$ one side & $y$ and $dy$ another side,

$\frac{dy}{y} = -7x^{2}\;dx$

On integrating both sides,

$\int \frac{dy}{y} = \int -7x^{2}\;dx$

$\Rightarrow$ $ln\;y = \frac{-7x^3}{3}+c$ , where ‘c’ is an arbitrary constant.

Since, It is given that $y(0)=\frac{3}{7}$

So, $c=ln\frac{3}{7}$

So, Solution of given differential equation will be :-

$ln\;y\;= \frac{-7x^3}{3}+ln\frac{3}{7}$

Now, when $x=1$,

$\Rightarrow$ $ln\;y\;=\;\frac{-7}{3}+ln\frac{3}{7}$

$\Rightarrow$ $ln\;y\;-ln\frac{3}{7} = \;\frac{-7}{3}$

$\Rightarrow$ $ln\;\frac{y}{(3/7)} = \frac{-7}{3}$

$\Rightarrow$ $ln\;\frac{7y}{3} = \frac{-7}{3}$

$\Rightarrow$ $\frac{7y}{3} = e^{ \frac{-7}{3}}$

$\Rightarrow$ $y=\frac{3}{7}e^{\frac{-7}{3}}$

So, Answer is $(C)$
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