Question

$$\frac{dy}{dx}={\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1}$$

Answer 1

We have,

$$\frac{dy}{dx}={\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1}$$

$$\Rightarrow dy=({\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1})dx$$

Integrating both sides, we get

$$\int dy=\int ({\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1})dx$$

$$\Rightarrow y=\int {\sin }^{3}x{\cos }^{4}xdx+\int x\sqrt{x+1}dx$$

$$\Rightarrow y=I_1+I_2.....\left(1\right)$$

Here,

$$I_1=\int {\sin }^{3}x{\cos }^{4}xdx$$

$$I_1=\int x\sqrt{x+1}dx$$
Now,
$$I_1=\int {\sin }^{3}x{\cos }^{4}xdx$$
$$=\int (1-{\cos }^{2}x){\cos }^{4}x\sin xdx$$
$$\text{Putting }t=\cos x,\text{ we get}$$
$$dt=-\sin xdx$$
$$\therefore I_1=-\int t^4(1-t^2)dt$$
$$=\int (t^6-t^4)dt$$
$$=\frac{t^7}{7}-\frac{t^5}{5}+C_1$$
$$=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+C_1$$
$$I_2=\int x\sqrt{x+1}dx$$
$$\text{Putting }{t}^2=x+1,\text{ we get}$$
$$2tdt=dx$$
$$\therefore I_2=2\int (t^2-1)t^{2}dt$$
$$=2\int (t^4-t^2)dt$$
$$=\frac{2t^5}{5}-\frac{2t^3}{3}+C_2$$
$$=\frac{2{(x+1)}^{\frac{5}{2}}}{5}-\frac{2{(x+1)}^{\frac{3}{2}}}{3}+C_2$$
$$\text{Putting the value of }{I}_1\text{ and }{I}_2\text{ in (1), we get}$$
$$y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+C_1+\frac{2{(x+1)}^{\frac{5}{2}}}{5}-\frac{2{(x+1)}^{\frac{3}{2}}}{3}+C_2$$
$$y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+\frac{2{(x+1)}^{\frac{5}{2}}}{5}-\frac{2{(x+1)}^{\frac{3}{2}}}{3}+C.............[\because C=C_1+C_2]$$
$$\text{Hence, }y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+\frac{2{(x+1)}^{\frac{5}{2}}}{5}-\frac{2{(x+1)}^{\frac{3}{2}}}{3}+C\text{ is the solution of the given differential equation.}$$