Question
The solution lof dydx−x tan(y−x)=1
Answer: Option A
:
A
Put y-x = z. Then dydx−1=dzdx⇒dydx=1+dzdx
Given dydx−xtan(y−x)=1⇒1+dzdx−xtanz=1⇒dzdx=xtanz⇒1tanzdz=xdx⇒∫cotzdz=∫xdx
⇒log|sinz|−logc=x22⇒log|sinzc|=x22=sinzc=ex22⇒sin(y−x)=cex22
∴ The solution is sin(y−x)=cex22, where c is arbitrary constant.
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:
A
Put y-x = z. Then dydx−1=dzdx⇒dydx=1+dzdx
Given dydx−xtan(y−x)=1⇒1+dzdx−xtanz=1⇒dzdx=xtanz⇒1tanzdz=xdx⇒∫cotzdz=∫xdx
⇒log|sinz|−logc=x22⇒log|sinzc|=x22=sinzc=ex22⇒sin(y−x)=cex22
∴ The solution is sin(y−x)=cex22, where c is arbitrary constant.
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