If the curve y-yx satisfies the differential equation dydx - 2xy1 - x2 - 0 and y0 - 1- then y1 is equal to

Dydx 2xy1 + x^2 nbsp;= 0 ⇒dyy = 2x dx1 + x^2 ⇒∫dyy = ∫2x dx1 + x^2 Let I = ∫2x dx1 + x^2 Assuming 1+x^2 = t ⇒ 2xdx= dt I = ∫dtt ⇒ I = ln |t| = ln |1+x^2 | ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

If the curve ...

Question

If the curve $y = y (x)$ satisfies the differential equation $\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0$ and $y (0) = 1,$ then $y (1)$ is equal to

1

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\frac{1}{2}

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Solution

The correct option is C $2$
$\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0$
$\Rightarrow \frac{d y}{y} = \frac{2 x d x}{1 + x^{2}} \Rightarrow \int \frac{d y}{y} = \int \frac{2 x d x}{1 + x^{2}}$

Let $I = \int \frac{2 x d x}{1 + x^{2}}$
Assuming $1 + x^{2} = t \Rightarrow 2 x d x = d t$
$I = \int \frac{d t}{t} \Rightarrow I = ln | t | = ln | 1 + x^{2} |$
Hence, the general solution is $ln | y | = ln | 1 + x^{2} | + ln | C |$
$\Rightarrow y = C (1 + x^{2})$
Putting $x = 0, y = 1$ , we get
$C = 1$
So, the required curve is,
$y = 1 + x^{2}$
$∴ y (1) = 2$

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If the curve y=y(x) satisfies the differential equation dydx−2xy1+x2=0 and y(0)=1, then y(1) is equal to

The correct option is C 2dydx−2xy1+x2=0 ⇒dyy=2xdx1+x2⇒∫dyy=∫2xdx1+x2 Let I=∫2xdx1+x2 Assuming 1+x2=t⇒2xdx=dt I=∫dtt⇒I=ln|t|=ln|1+x2| Hence, the general solution is ln|y|=ln|1+x2|+ln|C| ⇒y=C(1+x2) Putting x=0,y=1, we get C=1 So, the required curve is, y=1+x2 ∴y(1)=2

If the curve $y = y (x)$ satisfies the differential equation $\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0$ and $y (0) = 1,$ then $y (1)$ is equal to