The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $x+y \frac{dy}{dx}=\sec(x^2+y^2)$.Let $u = x^2+y^2$. Differentiating with respect to $x$,we get $\frac{du}{dx} =

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$\sin(x^2+y^2)=2x+c$

Vedclass · Answer

(A) Given the differential equation: $x+y \frac{dy}{dx}=\sec(x^2+y^2)$.Let $u = x^2+y^2$. Differentiating with respect to $x$,we get $\frac{du}{dx} = 2x + 2y \frac{dy}{dx}$.This implies $x + y \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx}$.Substituting this into the original equation,we have $\frac{1}{2} \frac{du}{dx} = \sec(u)$.Separating the variables,we get $\frac{du}{\sec(u)} = 2 dx$,which is $\cos(u) du = 2 dx$.Integrating both sides,we get $\int \cos(u) du = \int 2 dx$.This results in $\sin(u) = 2x + c$.Substituting back $u = x^2+y^2$,the general solution is $\sin(x^2+y^2) = 2x + c$.

The general solution of the differential equation $x+y \frac{dy}{dx}=\sec(x^2+y^2)$ is

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