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Exercise 4 · Q3

Q.Find the general solution of the differential equation: dydx=x+12−y\frac{dy}{dx}=\frac{x+1}{2-y}

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Cross-multiply to separate: (2−y) dy=(x+1) dx(2-y)\,dy=(x+1)\,dx, integrate, and simplify to x2+2x+y2−4y=Cx^2+2x+y^2-4y=C.

Variables-separable: bring all yy with dydy and all xx with dxdx, then integrate each side.

Steps

  1. Given:

dydx=x+12−y.\frac{dy}{dx}=\frac{x+1}{2-y}.

  1. Cross-multiply:

(2−y) dy=(x+1) dx.(2-y)\,dy=(x+1)\,dx.

  1. Integrate both sides:

∫(2−y) dy=∫(x+1) dx.\int(2-y)\,dy=\int(x+1)\,dx.

2y−y22=x22+x+C1.2y-\frac{y^2}{2}=\frac{x^2}{2}+x+C_1.

  1. Multiply through by 22:

4y−y2=x2+2x+2C1.4y-y^2=x^2+2x+2C_1.

  1. Rearrange (write C=−2C1C=-2C_1):

x2+2x+y2−4y=C.x^2+2x+y^2-4y=C.

✓Final answer

x2+2x+y2−4y=Cx^{2}+2x+y^{2}-4y=C.

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