CUET UG 2025 — Mathematics Calculus

Given:
$$\frac{dy}{dx} = e^{ax+by}$$dxdy​=eax+by

---

Using the exponential property $e^{A+B} = e^A \cdot e^B$eA+B=eA⋅eB:

$$\frac{dy}{dx} = e^{ax} \cdot e^{by}$$dxdy​=eax⋅eby

---

Separating variables:

$$\frac{dy}{e^{by}} = e^{ax} \, dx$$ebydy​=eaxdx

$$e^{-by} \, dy = e^{ax} \, dx$$e−bydy=eaxdx

---

Integrating both sides:

$$\int e^{-by} \, dy = \int e^{ax} \, dx$$∫e−bydy=∫eaxdx

For the left side, using $\displaystyle \int e^{ku} \, du = \frac{e^{ku}}{k}$∫ekudu=keku​ with $k = -b$k=−b:

$$\int e^{-by} \, dy = \frac{e^{-by}}{-b}$$∫e−bydy=−be−by​

$$= -\frac{e^{-by}}{b}$$=−be−by​

For the right side, with $k = a$k=a:

$$\int e^{ax} \, dx = \frac{e^{ax}}{a}$$∫eaxdx=aeax​

---

Combining results:

$$-\frac{e^{-by}}{b} = \frac{e^{ax}}{a} + C'$$−be−by​=aeax​+C′

Multiplying through by $-ab$−ab:

$$ae^{-by} = -be^{ax} - abC'$$ae−by=−beax−abC′

$$ae^{-by} + be^{ax} = -abC'$$ae−by+beax=−abC′

Since $-abC'$−abC′ is an arbitrary constant, replacing it with $C$C:

$$ae^{-by} + be^{ax} = C$$ae−by+beax=C