MHT CET 2023 14th May Morning Shift | Differential Equations Question 112 | Mathematics

The money invested in a company is compounded continuously. If ₹ 200 invested today becomes ₹ 400 in 6 years, then at the end of 33 years it will become ₹

$$1600 \sqrt{2}$$

$$3200 \sqrt{2}$$

$$12800 \sqrt{2}$$

$$6400 \sqrt{2}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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The differential equation of $$y=\mathrm{e}^x(\mathrm{a} \cos x+\mathrm{b} \sin x)$$ is

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}-y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

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If $$x d y=y(d x+y d y), y(1)=1, y(x)>0$$, then $$y(-3)$$ is

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

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The solution of $$(1+x y) y d x+(1-x y) x d y=0$$ is

$$\log \left(\frac{x}{y}\right)+\frac{1}{x y}=k$$, where $$k$$ is constant of integration

$$\log \left(\frac{x}{y}\right)=\frac{1}{x y}+k$$, where $$k$$ is constant of integration

$$\log \left(\frac{x}{y}\right)+{x y}=k$$, where $$k$$ is constant of integration

$$\log \left(\frac{x}{y}\right)={x y}+k$$, where $$k$$ is constant of integration

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