AP EAPCET 24 May 2025 Shift 1 Paper

Since, the slope of tangent to a curve at any point (x,y) is x+y.
∴‌‌‌

dy

dx

=x+y⇒‌

dy

dx

−y=x
This is a linear differentiation equation
So, P=−1 and Q=x
Now, IF=e∫P‌dx=e∫−dx=e−x
‌⇒‌‌e−x⋅‌

dy

dx

−e−xy=xe−x
‌⇒‌‌‌

dy

dx

(y⋅e−x)=xe−x
Integrating both sides, we get
y⋅e−x‌=∫xe−x‌dx
‌=−xe−x−∫−e−x‌dx
‌=−xe−x+(−e−x)+c
where C= constant of integration ⇒ye−x=−xe−x−e−x+c
⇒y=−x−1+cex, where c is arbitrary constant.