Question

Solve the differential equation $ydx - xdy + (1 + {x^2})dx + {x^2}\sin ydy = 0$

Answer 1

We divide all the terms by common term i.e. ${x^2}$ such that the term with dy is only of variable y. Cancel possible factors and then integrate all the terms. Use the formula of integration and quotient rule to write the terms in a simpler way. Use the integration of sine of angle as –cosine of angle. Shift all constants to the right hand side and write the solution.

• $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
• Quotient rule of derivative: If $y = \dfrac{f}{g} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{g \cdot f' - f \cdot g'}}{{{g^2}}}$ .

Complete step-by-step answer:
We are given the differential equation $ydx - xdy + (1 + {x^2})dx + {x^2}\sin ydy = 0$
Since the coefficient of dy is ${x^2}\sin y$ , we need to remove the term with variable x such that there is only a variable of y. We divide the complete equation by ${x^2}$ .
$\Rightarrow \dfrac{{ydx - xdy + (1 + {x^2})dx + {x^2}\sin ydy}}{{{x^2}}} = 0$
Separate the terms along with the denominator
$\Rightarrow \dfrac{{ydx - xdy}}{{{x^2}}} + \dfrac{{(1 + {x^2})dx}}{{{x^2}}} + \dfrac{{{x^2}\sin ydy}}{{{x^2}}} = 0$
Cancel same factors from numerator and denominator
$\Rightarrow \left( {\dfrac{{ydx - xdy}}{{{x^2}}}} \right) + \left( {\dfrac{1}{{{x^2}}} + 1} \right)dx + \sin ydy = 0$
We can bring negative sign from first term
$\Rightarrow - \left( {\dfrac{{xdy - ydx}}{{{x^2}}}} \right) + \left( {\dfrac{1}{{{x^2}}} + 1} \right)dx + \sin ydy = 0$
We know that according to quotient rule of differentiation $d\left( {\dfrac{y}{x}} \right) = \dfrac{{ydx - xdy}}{{{x^2}}}$
$\Rightarrow - d\left( {\dfrac{y}{x}} \right) + \left( {\dfrac{1}{{{x^2}}} + 1} \right)dx + \sin ydy = 0$
Now we integrate each term in left hand side of the equation
$\Rightarrow - \int {d\left( {\dfrac{y}{x}} \right)} + \int {\left( {\dfrac{1}{{{x^2}}} + 1} \right)} dx + \int {\sin ydy} = 0$
We can remove integration by derivative in first term
$\Rightarrow - \dfrac{y}{x} + \int {\left( {\dfrac{1}{{{x^2}}} + 1} \right)} dx + \int {\sin ydy} = 0$
Now we know formula of integration $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$ and substitute the value of $\int {\sin ydy} = - \cos y + C$
$\Rightarrow - \dfrac{y}{x} + \left( {\dfrac{{ - 1}}{x}} \right) + x + ( - \cos y) + C = 0$
$\Rightarrow - \dfrac{y}{x} - \dfrac{1}{x} + x - \cos y = C$
We take negative sign common
$\Rightarrow \dfrac{y}{x} + \dfrac{1}{x} - x + \cos y = C$

$\therefore$ Solution of the differential equation $ydx - xdy + (1 + {x^2})dx + {x^2}\sin ydy = 0$ is $\dfrac{y}{x} + \dfrac{1}{x} - x + \cos y = C$

Note:
Many students make the mistake of collecting the terms with dy and terms with dx separately and then try to write the value of $\dfrac{{dy}}{{dx}}$ by shifting the values to one side of the equation. Then many students try to integrate that but we cannot separate the terms of the variable of y and x so it’s difficult to integrate. Also, don’t write separate constants for each integration, instead write one constant that sums up and gives value as a final constant.