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NCERT Exemplar · Q20

Q.Solve: 2(y+3)−xydydx=02(y+3)-xy\frac{dy}{dx}=0, given that y(1)=−2y(1)=-2.

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Separable ODE. Integrating and applying y(1)=−2y(1)=-2 gives the implicit particular solution 2log⁡∣x∣+3log⁡∣y+3∣=y+22\log|x| + 3\log|y+3| = y + 2, i.e. x2(y+3)3=e y+2x^2(y+3)^3 = e^{\,y+2}.

Separate variables. From 2(y+3)=xydydx2(y+3) = xy\dfrac{dy}{dx}, divide by x(y+3)x(y+3):

yy+3 dy=2x dx.\frac{y}{y+3}\,dy = \frac{2}{x}\,dx.

Integrate. Writing yy+3=1−3y+3\dfrac{y}{y+3} = 1 - \dfrac{3}{y+3}:

∫ ⁣(1−3y+3)dy=∫2x dx⇒y−3log⁡∣y+3∣=2log⁡∣x∣+C.\int\!\left(1 - \frac{3}{y+3}\right)dy = \int \frac{2}{x}\,dx \quad\Rightarrow\quad y - 3\log|y+3| = 2\log|x| + C.

Apply the initial condition y(1)=−2y(1) = -2: then x=1, y+3=1x=1,\ y+3 = 1, so

−2−3log⁡1=2log⁡1+C⇒C=−2.-2 - 3\log 1 = 2\log 1 + C \quad\Rightarrow\quad C = -2.

Particular solution: …

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