CUET UG 2025 — Mathematics Calculus

The differential equation is $\frac{xdy}{dx} + 4y = x^3, (x \neq 0)$dxxdy​+4y=x3,(x=0)

Dividing the entire equation by $x$x:

$$\frac{dy}{dx} + \frac{4y}{x} = \frac{x^3}{x}$$dxdy​+x4y​=xx3​

$$\frac{dy}{dx} + \frac{4}{x}y = x^2$$dxdy​+x4​y=x2

This gives $P(x) = \frac{4}{x}$P(x)=x4​ and $Q(x) = x^2$Q(x)=x2

---

The integrating factor is:

$$I.F. = e^{\int P(x)dx}$$I.F.=e∫P(x)dx

$$I.F. = e^{\int \frac{4}{x}dx}$$I.F.=e∫x4​dx

$$I.F. = e^{4\ln x}$$I.F.=e4lnx

$$I.F. = e^{\ln x^4}$$I.F.=elnx4

$$I.F. = x^4$$I.F.=x4

---

The general solution is given by:

$$y \times I.F. = \int Q(x) \times I.F. \, dx + c$$y×I.F.=∫Q(x)×I.F.dx+c

$$y \times x^4 = \int x^2 \times x^4 \, dx + c$$y×x4=∫x2×x4dx+c

$$y \cdot x^4 = \int x^6 \, dx + c$$y⋅x4=∫x6dx+c

$$y \cdot x^4 = \frac{x^7}{7} + c$$y⋅x4=7x7​+c

---

Dividing both sides by $x^4$x4:

$$y = \frac{x^7}{7x^4} + \frac{c}{x^4}$$y=7x4x7​+x4c​

$$y = \frac{x^3}{7} + cx^{-4}$$y=7x3​+cx−4

Therefore, the general solution is $y = \frac{x^3}{7} + cx^{-4}$y=7x3​+cx−4 where $c$c is an arbitrary constant.