CBSE 2022 · Region 4 · Set 1 · Q12 · 4 marks
Find the particular solution of the differential equation \[x \frac{d y}{d x}+y+\frac{1}{1+x^{2}}=0 \text {, given that } y(1)=0 \text {. } \]Find the general solution of the differential equation \[x\left(y^{3}+x^{3}\right) d y=\left(2 y^{4}+5 x^{3} y\right) d x . \]
Find the particular solution of the differential equation \[x \frac{d y}{d x}+y+\frac{1}{1+x^{2}}=0 \text {, given that } y(1)=0 \text {. } \]
Find the general solution of the differential equation \[x\left(y^{3}+x^{3}\right) d y=\left(2 y^{4}+5 x^{3} y\right) d x . \]
Marking-scheme solution
(a)
Given differential equation can be written as \(\displaystyle \frac{d y}{d x}+\frac{1}{x} y=\frac{-1}{x\left(1+x^{2}\right)}\) I.F. \(\displaystyle =e^{\int \frac{1}{x} d x}=e^{\log x}=x\)
Solution is \(\displaystyle y \cdot x=\int \frac{-1}{1+x^{2}} d x+C \Rightarrow x y=-\tan ^{-1} x+C\)
Now \(\displaystyle y(1)=0 \Rightarrow C=\frac{\pi}{4}\) ∴ Particular solution is \(\displaystyle x y=\frac{\pi}{4}-\tan ^{-1} x\)
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplycase_studymedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.