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Question

# Solution of the differential equation dydx=sin(x+y)+cos(x+y) is:

A
log1+tanx+y2=y+c
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B
log2+secx+y2=x+c
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C
log|1+tan(x+y)|=y+c
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D
none of these
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Solution

## The correct option is C none of theseGiven,Differential equation as dydx=sin(x+y)+cos(x+y)−−−−>ALet (x+y)=tOn differentiating once⇒ddx(x+y)=dtdx⇒1+dydx=dtdxOn subsequent substitutions equation A becomes as follows.dtdx−1=sint+cost⇒dtdx=(sint+(1+cost))⇒dtdx=(2sint2cost2+(2cos2t2))⇒dtdx=2cos2t2(tant2+1)⇒2sec2t2(1+tant2)dt=dxon Integrating both sides⇒∫2sec2(t2)(1+tant2)=∫dx−−−−−−>Blet (1+tant2)=kon differentiating onceddt(1+tant2)=dkdt(0+sec2(t2))dt=dk(sec2(t2))dt=dk−−−−>COn substituting C in BA→∫dkk=∫dx⇒lnk=x+cBut k=(1+tant2)⇒ln(1+tant2)=x+cBut t=x+y⇒ln(1+tan(x+y)2)=x+c is the required solution of given diferential equation.

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