Question

The solution of the differential equation \[ 2x^{2}y\frac{dy}{dx}=\tan(x^{2}y^{2})-2xy^{2}, \] given \(y(1)=\sqrt{\frac{\pi}{2}}\) is:

• A. $\sin(x^{2}y^{2})=e^{x-1}$
• B. $\sin(x^{2}y^{2})=e^{2(x-1)}$
• C. $\cos\left(\frac{\pi}{2}+x^{2}y^{2}\right)+x=0$
• D. $\sin(x^{2}y^{2})=1$

Hint:

Whenever expressions like \(2x^{2}y\,dy+2xy^{2}\,dx\) appear together, check whether they form the exact differential of a combined term such as \(x^{2}y^{2}\).

Answer 1

The Correct Option is A

Given: \[ 2x^{2}y\frac{dy}{dx}=\tan(x^{2}y^{2})-2xy^{2} \] Rearrange: \[ 2x^{2}y\frac{dy}{dx}+2xy^{2}=\tan(x^{2}y^{2}) \] Multiply throughout by \(dx\): \[ 2x^{2}y\,dy+2xy^{2}\,dx=\tan(x^{2}y^{2})\,dx \] Notice that: \[ d(x^{2}y^{2})=2xy^{2}\,dx+2x^{2}y\,dy \] Let \[ u=x^{2}y^{2} \] Then: \[ du=\tan u\,dx \] So, \[ \frac{du}{\tan u}=dx \] \[ \cot u\,du=dx \] Integrating both sides: \[ \int \cot u\,du=\int dx \] \[ \ln|\sin u|=x+C \] Substituting back \(u=x^{2}y^{2}\): \[ \ln|\sin(x^{2}y^{2})|=x+C \] Using the condition: \[ y(1)=\sqrt{\frac{\pi}{2}} \] At \(x=1\): \[ x^{2}y^{2}=1^{2}\cdot\frac{\pi}{2}=\frac{\pi}{2} \] Hence, \[ \ln\left(\sin\frac{\pi}{2}\right)=1+C \] \[ \ln(1)=1+C \] \[ 0=1+C \] \[ C=-1 \] Therefore, \[ \ln|\sin(x^{2}y^{2})|=x-1 \] Exponentiating: \[ \sin(x^{2}y^{2})=e^{x-1} \] Hence, the correct answer is: \[ \boxed{\sin(x^{2}y^{2})=e^{x-1}} \]