Let x = exp be the solution of the differential equation 2 x 2 y ? d y - ( 1 - x y 2 ) d x = 0 , x 0 , y ( 2 ) = log e 2 . Then equals [2023]

Question

Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]

Smart Achievers · Accepted Answer

(3)

$α x = e^{x^{β} \cdot y^{γ}}$ and $2 x^{2} y \frac{d y}{d x} = 1 - x \cdot y^{2}$

Let $y^{2} = t \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$ $\Rightarrow x^{2} \frac{d t}{d x} = 1 - x t$

$\Rightarrow \frac{d t}{d x} + \frac{t}{x} = \frac{1}{x^{2}}$

$I.F. = e^{\log_{e} x} = x$

$t (x) = \int \frac{1}{x^{2}} \cdot x d x \Rightarrow y^{2} \cdot x = \log_{e} x + C$

$\Rightarrow 2 \log_{e} 2 = \log_{e} 2 + C \Rightarrow C = \log_{e} 2$

$Hence, x y^{2} = \log_{e} 2 x$ $∴$ $2 x = e^{x \cdot y^{2}}$

$Hence, α = 2, β = 1, γ = 2$ $∴$ $α + β - γ = 1$

Solution

Q.
Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]

1 0

2 - 1

3 1

4 3

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Solution

Q. Let αx=exp(xβyγ) be the solution of the differential equation 2x2y dy-(1-xy2)dx=0, x>0, y(2)=loge2. Then α+β-γ equals [2023]

1 0

2 - 1

3 1

4 3

Ans. (3) αx=exβ·yγ and 2x2ydydx=1-x·y2 Let y2=t⇒2ydydx=dtdx⇒x2dtdx=1-xt ⇒dtdx+tx=1x2 I.F.=elogex=x t(x)=∫1x2·xdx ⇒y2·x=logex+C ⇒2loge2=loge2+C ⇒ C=loge2 Hence, xy2=loge2x ∴ 2x=ex·y2 Hence, α=2, β=1, γ=2 ∴ α+β-γ=1

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Q.
Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]