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Mathematics · 2025 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2025 · Region 2 · Set 1 · Q26
(a) Solve the differential equation 2(y+3)-x y (d y)/(d x)=0; given y(1)=-2.
(a)
Solve the differential equation $\displaystyle 2(y+3)-x y \frac{d y}{d x}=0$; given $\displaystyle y(1)=-2$.
OR
(b)
Solve the following differential equation: \[\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} . \]
Marking-scheme solution
Given differential equation can be written as
\[\begin{aligned}
& \frac{y}{y+3} d y=\frac{2}{x} d x \\
& \Rightarrow \int\left(1-\frac{3}{y+3}\right) d y=2 \int \frac{1}{x} d x \\
& \Rightarrow y-3 \log |y+3|=2 \log |x|+C \\
& y=-2, \text { when } x=1 \Rightarrow C=-2
\end{aligned}
\]
Hence, the required particular solution is
\[\Rightarrow y-3 \log |y+3|=2 \log |x|-2
\]
Given differential equation can be written as
$\displaystyle \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}$, which is linear in y .
\[\text { I.F. }=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}
\]
The solution is given by
\[y\left(1+x^{2}\right)=\int 4 x^{2} d x
\]
$\displaystyle \Rightarrow y\left(1+x^{2}\right)=\frac{4}{3} x^{3}+C$
or $\displaystyle y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}+C \frac{1}{\left(1+x^{2}\right)}$, which is the required general solution
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.