Find the solution of differential equation d y d x = ^ 3 x 5 in ^ 4 x+x 2 x+1 :
Mathematics · JEE Main · NTA Exams — Integral Calculus
Find the solution of differential equation \(\frac{d y}{d x}=\cos ^{3} x \cdot 5 \operatorname{in}^{4} x+x \sqrt{2 x+1}\):
- \(y=\frac{\sin ^{5} x}{5}-\frac{\sin ^{7} x}{7}+\frac{(2 x+1)^{\frac{5}{2}}}{10}+\frac{(2 x+1)^{\frac{3}{2}}}{6}+c\)\(y=\frac{\sin ^{5} x}{5}-\frac{\sin ^{7} x}{7}+\frac{(2 x+1)^{\frac{5}{2}}}{10}-\frac{(2 x+1)^{\frac{3}{2}}}{6}+c\)\(y=\frac{\cos ^{5} x}{5}-\frac{\sin ^{7} x}{7}+\frac{(2 x+1)^{\frac{5}{2}}}{10}+\frac{(2 x+1)^{\frac{3}{2}}}{6}+c\)
- None of these
Answer
(B) y= ^ 5 x 5 - ^ 7 x 7 + (2 x+1)^ 5 2 10 - (2 x+1)^ 3 2 6 +c
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