Question

Find the particular solution of the differential equation $\dfrac{dy}{dx}=\dfrac{x(2\log x+1)}{\sin y+y\cos y}$ given that at x = 1 the value of y is $\dfrac{\pi }{2}$ .

Answer 1

Now we can see that in the given differential equation, x and y functions can be separated easily. Hence we will solve this using a variable separation method. So we will separate x and y variables and then integrate on both sides.

Complete step by step answer:
Now let us consider the given equation $\dfrac{dy}{dx}=\dfrac{x(2\log x+1)}{\sin y+y\cos y}$ .
Cross multiplying the equation we get $(\sin y+y\cos y)dy=x(2\log x+1)dx$
Now let us open the bracket.
$\sin ydy+y\cos ydy=(2x\log xdx+xdx)$
Now let us Integrate on Both sides.
$\int{\left( \sin ydy+y\cos ydy \right)}=\int{(2x\log xdx+xdx)}$
Now we know that $\int{f+g=\int{f}+\int{g}}$, using this property we get
$\int{\sin ydy}+\int{y\cos ydy}=\int{2x\log xdx}+\int{xdx}.....(1)$
Now consider each integrals separately
First let us consider the integral $\int{\sin ydy}$
$\int{\sin ydy=-\cos y........(2)}$
Hence now we have $\int{\sin ydy=-\cos y}$
Next let us consider the integral $\int{xdx}$
$\int{xdx=\dfrac{{{x}^{2}}}{2}}..........(3)$
Hence we have $\int{xdx=\dfrac{{{x}^{2}}}{2}}$
Now consider the integral $\int{y\cos ydy}$ . We will solve this by using integration by parts
Now we know that $\int{(u.v)=u\int{v}-\int{\left( u'\int{v} \right)}}$ . Hence if we take $u=y,v=\cos y$ we get.
$=-y\int{\cos ydy}-\int{1.\left( \int{\cos ydy} \right)dy}$
$\begin{aligned} & =y\sin y-\int{\sin y} \\\ & =y\sin y-(-\cos y) \\\ \end{aligned}$
Hence we get $\int{y\cos ydy}=y\sin y+\cos y..................(4)$
Now we consider the integral $\int{2x\log xdx}$
We will solve this too with the help of integration by parts.
Now we know that $\int{(u.v)=u\int{v}-\int{\left( u'\int{v} \right)}}$ . Hence if we take $u=\log x,v=x$ we get

& \int{x\log xdx}=\log x\int{xdx-\int{\left( \dfrac{1}{x} \right)\int{xdx}}}. \\\ & \int{x\log xdx}=\log x.\dfrac{{{x}^{2}}}{2}-\int{\left( \dfrac{1}{x}.\dfrac{{{x}^{2}}}{2} \right)}dx \\\ & \int{x\log xdx}=\log x.\dfrac{{{x}^{2}}}{2}-\int{\dfrac{x}{2}}dx \\\ & \int{x\log xdx}=\dfrac{{{x}^{2}}\log x}{2}-\dfrac{{{x}^{2}}}{4} \\\ & 2\int{x\log xdx}=\dfrac{{{x}^{2}}\log x}{1}-\dfrac{{{x}^{2}}}{2}...........(5) \\\ \end{aligned}$$ Hence we have $$\int{2x\log xdx}=\dfrac{{{x}^{2}}\log x}{1}-\dfrac{{{x}^{2}}}{2}$$ Now from equation (1), equation (2), equation (3), equation (4) and equation (5) we get. $\begin{aligned} & -\cos y+y\sin y+\cos y=\dfrac{{{x}^{2}}\log x}{1}-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{2}}}{2}+C \\\ & \Rightarrow y\sin y={{x}^{2}}\log x+C...........................(6) \\\ \end{aligned}$ Now we are given that at x = 1 the value of y = $\dfrac{\pi }{2}$ Hence we get $$\begin{aligned} & \dfrac{\pi }{2}\sin \left( \dfrac{\pi }{2} \right)={{1}^{2}}\log 1+C \\\ & \Rightarrow \dfrac{\pi }{2}(1)=0+C \\\ & \Rightarrow C=\dfrac{\pi }{2} \\\ \end{aligned}$$ Now we know hence substituting in equation (6) we get. $y\sin y={{x}^{2}}\log x+\dfrac{\pi }{2}$ **Note:** Here in the given differential equation we could easily separate the variables x and y. hence we could just solve by normal integration. Though while solving linear differential equations always check if the variables are properly separated if not go with the method of integrating factor.