Question
Solve the following differential equation
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
✓
Answer
We have,
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
$\Rightarrow\text{dy}=\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Integrating both sides, we get
$\Rightarrow\int\text{dy}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
$\Rightarrow\text{y}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{y}=\int\frac{1}{\text{t}^4}\ \text{dt}$
$=\frac{\text{t}^{-3}}{-3}+\text{C}$
$=\frac{-\sin^{-3}}{3}+\text{C}$
$=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$
hence, $\text{y}=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$ is the solution to the given differential equation.
$\sin^4\text{x}\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
$\Rightarrow\text{dy}=\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Integrating both sides, we get
$\Rightarrow\int\text{dy}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
$\Rightarrow\text{y}=\int\frac{\cos\text{x}}{\sin^4\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{y}=\int\frac{1}{\text{t}^4}\ \text{dt}$
$=\frac{\text{t}^{-3}}{-3}+\text{C}$
$=\frac{-\sin^{-3}}{3}+\text{C}$
$=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$
hence, $\text{y}=-\frac{1}{3}\text{cosec}^3\text{x}+\text{C}$ is the solution to the given differential equation.
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