AP EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}=\cos ^2(3 x+y)\) is \(\tan ^{-1}\left(\frac{\sqrt{3}}{2} \tan (3 x+y)\right)=f(x)\). Then, \(f(x)=\)
- A \(2 \sqrt{3}(x+C)\)
- B \(x+C\)
- C \(\frac{x+C}{2 \sqrt{3}}\)
- D \(\frac{\sqrt{3}}{2}(x+C)\)
Answer & Solution
Correct Answer
(A) \(2 \sqrt{3}(x+C)\)
Step-by-step Solution
Detailed explanation
Here, \(\frac{d y}{d x}=\cos ^2(3 x+y)\) On putting \(3 x+y=t\) \(3+\frac{d y}{d x}=\frac{d t}{d x}\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- The product of the perpendicular distances drawn from the origin to the pair of straight lines \(6 x^2-5 x y-6 y^2+x+5 y-1=0\) isAP EAMCET 2017 Hard
- If two of the lines represented by \(2 x^3+x^2 y+y^3=0\) are mutually perpendicular, then the slope of the third line isAP EAMCET 2017 Easy
- Let the equation of the tangent at a point \(\mathrm{P}\) on the parabola \(x^2-4 x-4 y+16=0\) be \(2 x-y-5=0\). If the equation of the normal drawn at \(\mathrm{P}\) to this parabola is \(\mathrm{ax}+\mathrm{y}+\mathrm{c}=0\), then \(\mathrm{ac}=\).0AP EAMCET 2023 Medium
- The area (is sq. units) bounded by the curves \(x=y^2\) and \(x=3-2 y^2\) isAP EAMCET 2024 Easy
- A random variable \(X\) has its range \(\{-1,0,1\}\). If its mean is 0.2 and \(P(X=0)=0.2\), then \(P(X=1)=\)AP EAMCET 2019 Medium
- If the roots of the equation \(x^3-13 x^2+\mathrm{K} x-27=0\) are in geometric progression then \(\mathrm{K}=\)AP EAMCET 2024 Medium
More PYQs from AP EAMCET
- In an acute angled triangle, \(\cot B \cot C+\cot A \cot C+\cot A \cot B\) is equal toAP EAMCET 2012 Easy
- Two masses \(90 \mathrm{~kg}\) and \(160 \mathrm{~kg}\) are separated by a distance of \(5 \mathrm{~m}\). The magnitude of intensity of the gravitational field at a point which is at a distance \(3 \mathrm{~m}\) from the \(90 \mathrm{~kg}\) mass and \(4 \mathrm{~m}\) from the \(160 \mathrm{~kg}\) mass is (Universal gravitational constant, \(\left.G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}\right)\)AP EAMCET 2019 Hard
- Identify the pair of molecules in which the central atom has same hybridizationAP EAMCET 2017 Easy
- Differentiation of \(\left(x^2-5 x+8\right) \times\left(x^3+7 x+9\right)\) can be done byAP EAMCET 2020 Easy
- \(\int_2^5 \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} d x=\)AP EAMCET 2022 Medium
- In \(\triangle \mathrm{ABC}\), if \(\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{C}}{2}=\frac{\mathrm{b}}{\mathrm{s}}\), then \(\sin \left(\frac{\mathrm{A}+\mathrm{C}}{3}\right)=\)AP EAMCET 2023 Hard