Question

Solve the differential equation: ${\sec }^{2}y\frac{dy}{dx}+\tan y=x^3.$

• A. $\tan y=x^3+3x^2+6x-6+c{e}^{-x}.$
• B. $\tan y=x^3-3x^2-6x-6+c{e}^{-x}.$
• C. $\tan y=x^3-3x^2+6x-6+c{e}^{-x}.$
• D. None of these.

Answer 1

The correct option is C $\tan y=x^3-3x^2+6x-6+c{e}^{-x}.$

${\sec }^{2}y\frac{dy}{dx}+\tan y=x^3$
Put $\tan y=v\Rightarrow {\sec }^{2}ydy=dv$
$\therefore \frac{dv}{dx}+v=x^3$ ...(1)
Here $P=1\Rightarrow \int Pdx=\int dx=x$
$\therefore I.F.=e^x$
Multiplying (1) by $I.F.$ we get
$e^x\frac{dv}{dx}+e^{x}v=e^x{x}^3$
Integrating both sides we get
$e^x.v=\int e^x{x}^{3}dx+c=e^x{x}^3-\int 3x^2{e}^{x}dx+c=e^x{x}^3-3x^2{e}^x+\int 6x{e}^{x}dx=e^x{x}^3-3x^2{e}^x+6x{e}^x-\int 6e^{x}dx=e^x{x}^3-3x^2{e}^x+6x{e}^x-6e^x\Rightarrow \tan y=x^3-3x^2+6x-6+c{e}^{-x}$