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NCERT Exemplar · Q28

Q.Find the general solution of dydx−3y=sin⁡2x\frac{dy}{dx}-3y=\sin 2x.

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The integrating factor e−3xe^{-3x} gives y=Ce3x−313sin⁡2x−213cos⁡2xy=Ce^{3x}-\frac{3}{13}\sin 2x-\frac{2}{13}\cos 2x.

Identify the form

dydx−3y=sin⁡2x\frac{dy}{dx}-3y=\sin 2x is linear, dydx+Py=Q\frac{dy}{dx}+Py=Q with P=−3P=-3 and Q=sin⁡2xQ=\sin 2x.

Integrating factor

I.F.=e∫(−3) dx=e−3x.\text{I.F.}=e^{\int(-3)\,dx}=e^{-3x}.

Multiplying the equation by it turns the left side into a single derivative:

ddx(e−3xy)=e−3xsin⁡2x.\frac{d}{dx}\big(e^{-3x}y\big)=e^{-3x}\sin 2x.

Integrate the right side

With the standard result ∫eaxsin⁡bx dx=eaxa2+b2(asin⁡bx−bcos⁡bx)\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx) and a=−3, b=2a=-3,\ b=2 (a2+b2=13a^2+b^2=13):

e−3xy=e−3x13(−3sin⁡2x−2cos⁡2x)+C.e^{-3x}y=\frac{e^{-3x}}{13}(-3\sin 2x-2\cos 2x)+C. …

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